I85PREP
⌘ K
Abitur Mathematics study materials
Abitur Mathematics study resources

Abitur Mathematics practice questions

One question at a time. Build your Abitur Mathematics understanding.

First 50 of 200 questions. Each one lists the choices, the correct answer, and an explanation. The checker at the top of this page saves progress in this browser when it is available.

Continue in the practice tool above to load more questions, 50 at a time.

Question 1Analysis: functions, derivatives, integrals, curve sketching

What is the derivative of f(x) = x³ − 4x + 7 with respect to x?

  1. x² − 4
  2. 3x³ − 4
  3. 3x² − 4 Correct answer.
  4. 3x² − 4x

Explanation

The power rule gives d/dx(x³) = 3x² and d/dx(−4x) = −4, while the constant 7 differentiates to 0. Adding these terms yields f′(x) = 3x² − 4.

Question 2Analysis: functions, derivatives, integrals, curve sketching

A function f is defined on the closed interval [1, 5] and is continuous there. Which statement is guaranteed by the Extreme Value Theorem?

  1. f attains both a maximum and a minimum value on [1, 5] Correct answer.
  2. f has a zero in (1, 5)
  3. f is strictly increasing on [1, 5]
  4. f is differentiable at every point of (1, 5)

Explanation

The Extreme Value Theorem states that a continuous function on a closed bounded interval attains a global maximum and a global minimum on that interval. Continuity alone does not force monotonicity, differentiability, or the existence of a zero.

Question 3Analysis: functions, derivatives, integrals, curve sketching

Evaluate the definite integral ∫₀² (3x² + 1) dx.

  1. 12
  2. 10 Correct answer.
  3. 9
  4. 6

Explanation

An antiderivative is x³ + x. Evaluating at the bounds gives (2³ + 2) − (0 + 0) = 8 + 2 = 10, so the integral equals 10.

Question 4Analysis: functions, derivatives, integrals, curve sketching

The graph of a function f has a horizontal tangent at x = 2 and f changes from increasing to decreasing there. What does this indicate about f at x = 2?

  1. A local maximum Correct answer.
  2. A vertical asymptote
  3. A point of inflection with horizontal tangent
  4. A local minimum

Explanation

A sign change of f′ from positive to negative means the function turns from rising to falling, producing a local maximum. The horizontal tangent confirms f′(2) = 0, consistent with a stationary extremum.

Question 5Analysis: functions, derivatives, integrals, curve sketching

Which expression gives the derivative of g(x) = (2x + 1)⁵?

  1. 5(2x + 1)⁴
  2. 10(2x + 1)⁴ Correct answer.
  3. 10(2x + 1)⁵
  4. 2(2x + 1)⁴

Explanation

The chain rule multiplies the outer derivative 5(2x + 1)⁴ by the inner derivative 2. This gives g′(x) = 10(2x + 1)⁴.

Question 6Analysis: functions, derivatives, integrals, curve sketching

What is the value of ∫₁ᵉ (1/x) dx, where e is the Euler number?

  1. e
  2. e − 1
  3. 1/e
  4. 1 Correct answer.

Explanation

An antiderivative of 1/x is ln x. Evaluating gives ln e − ln 1 = 1 − 0 = 1, so the integral equals 1.

Question 7Analysis: functions, derivatives, integrals, curve sketching

For the function f(x) = x² − 6x + 5, at which x-value does the graph have its vertex (turning point)?

  1. x = 6
  2. x = −3
  3. x = 3 Correct answer.
  4. x = 5

Explanation

Setting f′(x) = 2x − 6 = 0 gives x = 3, and since the leading coefficient is positive this is a minimum. Alternatively, the vertex of ax² + bx + c lies at x = −b/(2a) = 3.

Question 8Analysis: functions, derivatives, integrals, curve sketching

Which statement about the graph of f(x) = x³ is correct on the whole real line?

  1. It has a local maximum at the origin
  2. It is strictly increasing and has a saddle point at the origin Correct answer.
  3. It has two local extrema
  4. It is bounded above

Explanation

Here f′(x) = 3x² ≥ 0, vanishing only at x = 0, so f is strictly increasing everywhere. At the origin the tangent is horizontal but no sign change of f′ occurs, giving a saddle (terrace) point.

Question 9Analysis: functions, derivatives, integrals, curve sketching

The function f has f(1) = 3 and f′(x) > 0 for all x in [1, 4]. Which conclusion is valid?

  1. f(4) < 3
  2. f(4) = 3
  3. f(4) > 3 Correct answer.
  4. f has a local maximum at x = 1

Explanation

If f′ > 0 on an interval, f is strictly increasing there, so moving from x = 1 to x = 4 raises the value. Hence f(4) > f(1) = 3.

Question 10Analysis: functions, derivatives, integrals, curve sketching

Compute the derivative of h(x) = x·ln x for x > 0.

  1. x/ln x
  2. ln x + 1 Correct answer.
  3. 1/x
  4. ln x

Explanation

The product rule gives h′(x) = 1·ln x + x·(1/x) = ln x + 1. The second term simplifies to 1.

Question 11Analysis: functions, derivatives, integrals, curve sketching

Determine the value of ∫₀^π sin x dx.

  1. 0
  2. 2 Correct answer.
  3. π
  4. 1

Explanation

An antiderivative of sin x is −cos x. Evaluating gives (−cos π) − (−cos 0) = 1 + 1 = 2.

Question 12Analysis: functions, derivatives, integrals, curve sketching

The graph of f has f″(x) < 0 on (0, 3) and f″(x) > 0 on (3, 6). What happens at x = 3?

  1. The function has a discontinuity
  2. There is a local maximum
  3. There is a local minimum
  4. The concavity changes, so there is an inflection point Correct answer.

Explanation

A sign change of the second derivative indicates a change from concave down to concave up, which defines an inflection point. Whether it is also an extremum depends on f′, which is not given.

Question 13Analysis: functions, derivatives, integrals, curve sketching

Find the equation of the tangent line to f(x) = x² at x = 3.

  1. y = 3x − 9
  2. y = 6x + 9
  3. y = 6x − 9 Correct answer.
  4. y = 9x − 18

Explanation

We have f(3) = 9 and f′(x) = 2x, so the slope is f′(3) = 6. Point-slope form gives y − 9 = 6(x − 3), that is y = 6x − 9.

Question 14Analysis: functions, derivatives, integrals, curve sketching

The function F is defined by F(x) = ∫₀^x (t² + 1) dt. What is F′(x)?

  1. 2x
  2. x² + 1 Correct answer.
  3. x²
  4. x³/3 + x

Explanation

By the Fundamental Theorem of Calculus, differentiating an accumulation function with variable upper limit returns the integrand evaluated at that limit. Hence F′(x) = x² + 1.

Question 15Analysis: functions, derivatives, integrals, curve sketching

A function satisfies f′(x) = 0 at x = 0 and x = 4, with f′(x) > 0 on (−∞, 0), f′(x) < 0 on (0, 4), and f′(x) > 0 on (4, ∞). What are the extrema?

  1. Local maxima at both x = 0 and x = 4
  2. Local minimum at x = 0 and local maximum at x = 4
  3. No extrema at all
  4. Local maximum at x = 0 and local minimum at x = 4 Correct answer.

Explanation

At x = 0 the derivative changes from positive to negative, giving a local maximum. At x = 4 it changes from negative to positive, giving a local minimum.

Question 16Analysis: functions, derivatives, integrals, curve sketching

Evaluate ∫₁² (2x − 1/x) dx.

  1. 1 − ln 2
  2. 3 − ln 2 Correct answer.
  3. 3 + ln 2
  4. 2 − ln 2

Explanation

An antiderivative is x² − ln x. Evaluating gives (4 − ln 2) − (1 − 0) = 3 − ln 2.

Question 17Analysis: functions, derivatives, integrals, curve sketching

Which condition guarantees that a differentiable function f has a point c in (a, b) with f′(c) = 0?

  1. f(a) = f(b), by Rolle's theorem Correct answer.
  2. f(a) = 0 and f(b) = 0 only at the endpoints
  3. f″(x) > 0 on (a, b)
  4. f is increasing on [a, b]

Explanation

Rolle's theorem requires continuity on [a, b], differentiability on (a, b), and equal endpoint values. These hypotheses then guarantee an interior point with horizontal tangent.

Question 18Analysis: functions, derivatives, integrals, curve sketching

The graph of f(x) = (x² − 1)/(x² + 1) is considered. Which statement about its asymptotes is correct?

  1. It has the horizontal asymptote y = 0 only
  2. It has the horizontal asymptote y = 1 and no vertical asymptotes Correct answer.
  3. It has an oblique asymptote y = x
  4. It has vertical asymptotes at x = 1 and x = −1

Explanation

As x → ±∞ the ratio tends to 1 because numerator and denominator have the same degree with leading coefficient 1. The denominator x² + 1 is never zero, so no vertical asymptote exists.

Question 19Analysis: functions, derivatives, integrals, curve sketching

What is the derivative of f(x) = e^(3x) · sin x?

  1. e^(3x)(3 sin x + cos x) Correct answer.
  2. 3e^(3x) cos x
  3. e^(3x)(sin x + 3 cos x)
  4. 3e^(3x) sin x

Explanation

The product rule combined with the chain rule gives f′(x) = 3e^(3x) sin x + e^(3x) cos x. Factoring e^(3x) yields e^(3x)(3 sin x + cos x).

Question 20Analysis: functions, derivatives, integrals, curve sketching

The derivative of f is f′(x) = (x − 2)²(x + 1). At which x-value does f have a local extremum?

  1. x = −1 only Correct answer.
  2. Neither value
  3. x = −1 and x = 2
  4. x = 2 only

Explanation

The factor (x − 2)² is non-negative, so f′ changes sign only at x = −1, from negative to positive, producing a minimum. At x = 2 the derivative touches zero without changing sign, so no extremum occurs.

Question 21Analysis: functions, derivatives, integrals, curve sketching

Let f(x) = x³ − 3x². Determine the x-coordinates of the inflection point of its graph.

  1. x = −1
  2. x = 1 Correct answer.
  3. x = 2
  4. x = 0 and x = 2

Explanation

We compute f′(x) = 3x² − 6x and f″(x) = 6x − 6. Setting f″ = 0 gives x = 1, and since f″ changes sign there, it is an inflection point.

Question 22Analysis: functions, derivatives, integrals, curve sketching

A function f is continuous on [0, 4] with ∫₀⁴ f(x) dx = 10. What is the average value of f on [0, 4]?

  1. 0.4
  2. 40
  3. 2.5 Correct answer.
  4. 10

Explanation

The average value of a continuous function on [a, b] is (1/(b − a))∫ₐᵇ f(x) dx. Here this equals 10/4 = 2.5.

Question 23Analysis: functions, derivatives, integrals, curve sketching

Consider f(x) = x⁴ − 4x³. How many local extrema does the graph have, and of what type?

  1. One local minimum at x = 3 Correct answer.
  2. Two local extrema, at x = 0 and x = 3
  3. One local maximum at x = 3
  4. No local extrema

Explanation

Here f′(x) = 4x³ − 12x² = 4x²(x − 3). The factor x² does not change sign at x = 0, so there is a saddle point there. At x = 3 the derivative changes from negative to positive, giving a single local minimum.

Question 24Analysis: functions, derivatives, integrals, curve sketching

The function f satisfies f′(x) = f(x) for all x and f(0) = 2. Which expression describes f?

  1. f(x) = 2e^x Correct answer.
  2. f(x) = x² + 2
  3. f(x) = 2 + x
  4. f(x) = e^(2x)

Explanation

The only solutions of y′ = y are y = C·e^x. The initial condition f(0) = 2 forces C = 2, so f(x) = 2e^x.

Question 25Analysis: functions, derivatives, integrals, curve sketching

The region bounded by the graph of f(x) = x², the x-axis, and the line x = 3 is rotated about the x-axis. Which integral gives the volume of the solid of revolution?

  1. π∫₀³ x⁴ dx Correct answer.
  2. π∫₀³ x² dx
  3. 2π∫₀³ x³ dx
  4. π∫₀³ x dx

Explanation

The disc method uses cross-sectional area π(f(x))² = πx⁴, integrated from 0 to 3. The volume is therefore π∫₀³ x⁴ dx, which evaluates to 243π/5.

Question 26Analysis: functions, derivatives, integrals, curve sketching

What is the domain of the function f(x) = ln(x − 2)?

  1. (2, ∞) Correct answer.
  2. ℝ \ {2}
  3. [2, ∞)
  4. (−∞, 2)

Explanation

The natural logarithm is defined only for positive arguments, so x − 2 > 0, which gives x > 2. The endpoint 2 is excluded because ln(0) is undefined.

Question 27Analysis: functions, derivatives, integrals, curve sketching

What is the limit of f(x) = (3x² + 2x − 1)/(x² − 4) as x → ∞?

  1. −1/4
  2. 0
  3. ∞
  4. 3 Correct answer.

Explanation

For a rational function, the limit at infinity is the ratio of the leading coefficients when numerator and denominator have the same degree. Here both are quadratic with leading coefficients 3 and 1, so the limit is 3.

Question 28Analysis: functions, derivatives, integrals, curve sketching

Which expression gives the derivative of f(x) = (2x + 1)^5?

  1. 2(2x + 1)^4
  2. (2x + 1)^4
  3. 10(2x + 1)^4 Correct answer.
  4. 5(2x + 1)^4

Explanation

The chain rule gives f′(x) = 5(2x + 1)^4 · 2, because the inner derivative of 2x + 1 is 2. Multiplying yields 10(2x + 1)^4.

Question 29Analysis: functions, derivatives, integrals, curve sketching

What is the value of the definite integral ∫₀¹ (4x³ − 2x) dx?

  1. −1
  2. 1
  3. 2
  4. 0 Correct answer.

Explanation

An antiderivative is x⁴ − x². Evaluating at 1 gives 1 − 1 = 0, and at 0 gives 0, so the integral equals 0 − 0 = 0.

Question 30Analysis: functions, derivatives, integrals, curve sketching

The graph of a function f has a horizontal asymptote y = 2 as x → ∞. Which statement about f is necessarily true?

  1. The values f(x) get arbitrarily close to 2 for sufficiently large x. Correct answer.
  2. The graph intersects the line y = 2 at infinitely many points.
  3. The derivative f′(x) tends to 0 as x → ∞.
  4. The function f is bounded on its entire domain.

Explanation

By definition, lim_{x→∞} f(x) = 2 means that for large x the function values approach 2 arbitrarily closely. This does not require the graph to meet the line y = 2, nor does it control the derivative or the behavior near other points.

Question 31Analysis: functions, derivatives, integrals, curve sketching

What is the derivative of f(x) = sin(3x) · cos x?

  1. 3 cos(3x) cos x − sin(3x) sin x Correct answer.
  2. −3 sin(3x) sin x
  3. 3 cos(3x) cos x + sin(3x) sin x
  4. 3 cos(3x) − sin x

Explanation

The product rule gives f′ = (sin(3x))′ · cos x + sin(3x) · (cos x)′. Since (sin(3x))′ = 3 cos(3x) and (cos x)′ = −sin x, the result is 3 cos(3x) cos x − sin(3x) sin x.

Question 32Analysis: functions, derivatives, integrals, curve sketching

The function f is defined by f(x) = x² − 4x + 1. On which interval is f strictly decreasing?

  1. (2, ∞)
  2. (−∞, 2) Correct answer.
  3. (0, 4)
  4. (−∞, 4)

Explanation

The derivative is f′(x) = 2x − 4, which is negative exactly when x < 2. A negative derivative on an interval means the function is strictly decreasing there.

Question 33Analysis: functions, derivatives, integrals, curve sketching

Which integral represents the area enclosed by the graph of f(x) = x², the x-axis, and the vertical lines x = 1 and x = 3?

  1. ∫₀¹ x² dx
  2. ∫₁³ x² dx Correct answer.
  3. ∫₀³ x² dx
  4. ∫₁³ 2x dx

Explanation

The area under a nonnegative curve between x = a and x = b is the definite integral of the function from a to b. Here the bounds are 1 and 3 and the function is x².

Question 34Analysis: functions, derivatives, integrals, curve sketching

Consider the function f(x) = x³ − 6x² + 9x. What are the x-coordinates of its local extrema?

  1. x = 2 only
  2. x = 1 and x = 3 Correct answer.
  3. x = 0 and x = 2
  4. x = −1 and x = 3

Explanation

Setting f′(x) = 3x² − 12x + 9 = 3(x − 1)(x − 3) equal to zero gives x = 1 and x = 3. The derivative changes sign at both, so both are local extrema.

Question 35Analysis: functions, derivatives, integrals, curve sketching

The function f has derivative f′(x) = (x − 1)²(x + 2). Which statement about the graph of f is correct?

  1. f has a local maximum at x = −2 and a local minimum at x = 1.
  2. f has no local extrema.
  3. f has a local minimum at x = −2 and no extremum at x = 1. Correct answer.
  4. f has local maxima at x = 1 and x = −2.

Explanation

The factor (x − 1)² is nonnegative and does not change sign at x = 1, so there is no extremum there. The factor x + 2 changes from negative to positive at x = −2, so f′ changes from negative to positive, giving a local minimum.

Question 36Analysis: functions, derivatives, integrals, curve sketching

What is the value of ∫₀^(π/2) x · sin x dx?

  1. π/2 − 1
  2. π/2
  3. 0
  4. 1 Correct answer.

Explanation

Integration by parts with u = x and dv = sin x dx gives [−x cos x]₀^(π/2) + ∫₀^(π/2) cos x dx. The boundary term is 0, and the remaining integral equals [sin x]₀^(π/2) = 1.

Question 37Analysis: functions, derivatives, integrals, curve sketching

The function F is defined by F(x) = ∫₁^x (1/t) dt for x > 0. Which expression equals F(x)?

  1. ln x − 1
  2. e^x
  3. ln x Correct answer.
  4. 1/x

Explanation

An antiderivative of 1/t is ln t, so F(x) = [ln t]₁^x = ln x − ln 1 = ln x. The lower limit 1 makes the constant vanish.

Question 38Analysis: functions, derivatives, integrals, curve sketching

The graph of f(x) = (2x − 1)/(x + 3) has which pair of asymptotes?

  1. Vertical x = −3 and horizontal y = −1/3
  2. Vertical x = −3 and horizontal y = 2 Correct answer.
  3. Vertical x = 1/2 and horizontal y = 0
  4. Vertical x = 3 and horizontal y = 2

Explanation

A vertical asymptote occurs where the denominator is zero and the numerator is not, so x = −3. Since numerator and denominator have the same degree, the horizontal asymptote is the ratio of leading coefficients, y = 2.

Question 39Analysis: functions, derivatives, integrals, curve sketching

The function f is continuous on [0, 2] and satisfies f(0) = −1 and f(2) = 3. Which conclusion follows from the intermediate value theorem?

  1. f is increasing on [0, 2].
  2. f′(c) = 0 for some c in (0, 2).
  3. There exists c in (0, 2) with f(c) = 1. Correct answer.
  4. The maximum value of f on [0, 2] is 3.

Explanation

The intermediate value theorem states that a continuous function on an interval attains every value between its endpoint values. Since 1 lies between −1 and 3, some c in (0, 2) has f(c) = 1.

Question 40Analysis: functions, derivatives, integrals, curve sketching

What is the average rate of change of f(x) = x³ on the interval [1, 3]?

  1. 26
  2. 14
  3. 9
  4. 13 Correct answer.

Explanation

The average rate of change is (f(3) − f(1))/(3 − 1) = (27 − 1)/2 = 13. This equals the slope of the secant line through the points (1, 1) and (3, 27).

Question 41Analysis: functions, derivatives, integrals, curve sketching

The graph of f has a point of inflection at x = 2, and f′(2) = 0. Which statement is necessarily true?

  1. f″ changes sign at x = 2. Correct answer.
  2. f has a local maximum at x = 2.
  3. f has a local minimum at x = 2.
  4. f‴(2) = 0.

Explanation

A point of inflection occurs where the concavity of the graph changes, which means f″ changes sign there. The condition f′(2) = 0 does not force an extremum; it may be a saddle-like stationary inflection point.

Question 42Analysis: functions, derivatives, integrals, curve sketching

The function f satisfies f′(x) = 3x² + 1 and f(1) = 4. What is f(x)?

  1. x³ + x + 4
  2. x³ + x + 2 Correct answer.
  3. 6x
  4. x³ + x + 3

Explanation

Integrating f′ gives f(x) = x³ + x + C. Substituting x = 1 and f(1) = 4 gives 1 + 1 + C = 4, so C = 2.

Question 43Analysis: functions, derivatives, integrals, curve sketching

The function f is defined by f(x) = x·e^(−x). Which statement about its graph is correct?

  1. It has no extrema and no asymptotes.
  2. It has a local maximum at x = 1 and a horizontal asymptote y = 0 as x → ∞. Correct answer.
  3. It has a local maximum at x = 1 and a vertical asymptote x = 0.
  4. It has a local minimum at x = 1 and a horizontal asymptote y = 0 as x → ∞.

Explanation

The derivative f′(x) = e^(−x)(1 − x) is positive for x < 1 and negative for x > 1, so f has a local maximum at x = 1. As x → ∞, e^(−x) dominates and f(x) → 0, giving a horizontal asymptote y = 0.

Question 44Analysis: functions, derivatives, integrals, curve sketching

What is the value of ∫₁^e (ln x)/x dx?

  1. 1/e
  2. e − 1
  3. 1
  4. 1/2 Correct answer.

Explanation

Substituting u = ln x gives du = (1/x) dx, so the integral becomes ∫₀¹ u du = [u²/2]₀¹ = 1/2. The limits change because ln 1 = 0 and ln e = 1.

Question 45Analysis: functions, derivatives, integrals, curve sketching

The function f is differentiable on ℝ and satisfies f′(x) = 0 for all x in an interval (a, b). Which statement about f on (a, b) is necessarily true?

  1. f is strictly increasing on (a, b).
  2. f is constant on (a, b). Correct answer.
  3. f is linear but not constant on (a, b).
  4. f has a local extremum at every point of (a, b).

Explanation

By the mean value theorem, if f′ = 0 on an interval, then for any two points in that interval the difference quotient is zero, so the function values are equal. Hence f is constant on the interval.

Question 46Analysis: functions, derivatives, integrals, curve sketching

The region bounded by the graph of f(x) = √x, the x-axis, and the line x = 4 is rotated about the x-axis. Which integral gives the volume of the solid?

  1. 2π ∫₀⁴ x√x dx
  2. π ∫₀⁴ x dx Correct answer.
  3. π ∫₀⁴ √x dx
  4. π ∫₀⁴ x² dx

Explanation

Using the disk method, the volume is π ∫₀⁴ [f(x)]² dx. Since (√x)² = x, the integral becomes π ∫₀⁴ x dx, which evaluates to 8π.

Question 47Analysis: functions, derivatives, integrals, curve sketching

The function f is defined by f(x) = x⁴ − 2x². How many points of inflection does its graph have?

  1. 3
  2. 0
  3. 1
  4. 2 Correct answer.

Explanation

The second derivative is f″(x) = 12x² − 4 = 4(3x² − 1), which vanishes at x = ±1/√3. Because f″ changes sign at each of these two points, the graph has two inflection points.

Question 48Analysis: functions, derivatives, integrals, curve sketching

The function f is continuous on [−1, 2] and satisfies ∫_{−1}^2 f(x) dx = 6. What is the average value of f on [−1, 2]?

  1. 1/2
  2. 2 Correct answer.
  3. 6
  4. 3

Explanation

The average value of a continuous function on [a, b] is (1/(b − a)) ∫_a^b f(x) dx. Here b − a = 3, so the average is 6/3 = 2.

Question 49Analysis: functions, derivatives, integrals, curve sketching

The function f is defined by f(x) = (x² − 1)/(x² − 4). Which statement about its graph is correct?

  1. It has vertical asymptotes x = 2 and x = −2, and a horizontal asymptote y = 0.
  2. It has vertical asymptotes x = 1 and x = −1, and a horizontal asymptote y = 1.
  3. It has vertical asymptotes x = 2 and x = −2, and a horizontal asymptote y = 1. Correct answer.
  4. It has no vertical asymptotes and a horizontal asymptote y = 1.

Explanation

The denominator vanishes at x = ±2 while the numerator does not, giving vertical asymptotes. Since numerator and denominator have the same degree, the horizontal asymptote is the ratio of leading coefficients, y = 1.

Question 50Analysis: functions, derivatives, integrals, curve sketching

The function f is defined by f(x) = x²·e^x. At which x-values does its graph have a point of inflection?

  1. x = −2 ± √2 Correct answer.
  2. x = −2 only
  3. x = 0 and x = −2
  4. x = −1 only

Explanation

Differentiating twice gives f″(x) = e^x(x² + 4x + 2). Setting x² + 4x + 2 = 0 and solving with the quadratic formula yields x = −2 ± √2. Since the quadratic changes sign at each simple root, both are inflection points.