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A square root is real only when its argument is non-negative, so we need x − 3 ≥ 0, giving x ≥ 3. The endpoint 3 is included because √0 = 0 is defined.
To invert, write y = 2x + 5 and solve for x: x = (y − 5)/2. Swapping the variable name gives f⁻¹(x) = (x − 5)/2.
Complete the square: x² − 4x + 7 = (x − 2)² + 3. Since a square is at least 0, the minimum value is 3, attained at x = 2, so the range is y ≥ 3.
The composition (f ∘ g)(x) means f(g(x)). Substituting g(x) = x + 1 into f gives (x + 1)², not x² + 1.
A function is odd when f(−x) = −f(x) for all x. For x³ + x we get (−x)³ + (−x) = −x³ − x = −(x³ + x), so it is odd.
A vertical asymptote occurs where the denominator vanishes and the numerator is nonzero. Here x − 2 = 0 gives x = 2, where the function is undefined.
The tangent function has period π, and the factor 2 inside the argument compresses the graph horizontally. The period of tan(kx) is π/|k|, so here it is π/2.
The transformation y = f(x − 2) shifts the graph 2 units to the right. A point (a, b) moves to (a + 2, b), so (3, 5) becomes (5, 5).
The expression is undefined only where the denominator is zero, that is x − 4 = 0. Every other real number is allowed, so the domain is all real x except 4.
Completing the square gives (x − 3)² − 4, with vertex at x = 3. Since x = 3 is included in the restricted domain, the minimum value is −4 and the range is y ≥ −4.
The composition (g ∘ f)(2) means g(f(2)). First f(2) = 5, then g(5) = 25 − 3 = 22.
Reflecting in the y-axis replaces x with −x, giving y = f(−x). Shifting 3 units down then subtracts 3 from the whole expression, giving y = f(−x) − 3.
Write y = (x + 1)/(x − 2) and solve: y(x − 2) = x + 1, so x(y − 1) = 2y + 1 and x = (2y + 1)/(y − 1). Swapping names gives f⁻¹(x) = (2x + 1)/(x − 1).
A quadratic has an inverse only when restricted to one side of its axis of symmetry. Here the axis is x = 1, so restricting to x ≥ 1 makes f one-to-one and hence invertible.
The expression |x − h| + k has a vertex at (h, k). Here h = 2 and k = 1, so the vertex is (2, 1).
The square root √(x + 4) takes all values y ≥ 0, so 3 − √(x + 4) takes all values at most 3. The maximum value 3 occurs when x = −4, giving range y ≤ 3.
For an even function f(−x) = f(x), so f(−3) = 7. For an odd function g(−x) = −g(x), so g(−3) = −7. The sum is 7 + (−7) = 0.
Replacing x by x/2 stretches the graph horizontally by factor 2, since each point moves twice as far from the y-axis. Multiplying the whole function by 2 stretches it vertically by factor 2.
On this interval the function is increasing. At x = −1 the value is −1 + 3 = 2, and at x = 1 the value is 1 − 3 = −2, so the range is [−2, 2].
Adding 5 outside the function shifts the whole graph 5 units up, so every horizontal line moves up by 5. The asymptote y = 2 becomes y = 7.
Substituting gives f(g(x)) = 2·((x − 3)/2) + 3 = (x − 3) + 3 = x, and similarly g(f(x)) = x. Since each undoes the other, they are inverse functions.
An absolute value equation splits into two cases: 2x − 1 = 5 gives x = 3, and 2x − 1 = −5 gives x = −2. Both values satisfy the original equation, giving the set {3, −2}.
The denominator factors as (x − 3)(x + 3), so it vanishes at x = 3 and x = −3, where the numerator is nonzero, producing two vertical asymptotes. Since the denominator has higher degree than the numerator, the horizontal asymptote is y = 0.
Completing the square gives f(x) = (x + 2)² + 3. Since x ≤ −2, we have x + 2 ≤ 0, so x + 2 = −√(y − 3). Thus f⁻¹(x) = −2 − √(x − 3), with domain x ≥ 3 matching the range of f.
Write f(2 − x) = f(−(x − 2)). Replacing x with −x reflects in the y-axis, and replacing x with x − 2 then shifts 2 units right. Order matters when combining these transformations.
Completing the square gives f(x) = (x − 2)² − 3, so the vertex is at x = 2, which lies inside the interval. The minimum value is therefore −3. Checking the endpoints, f(0) = 1 and f(5) = 6, so the maximum value is 6 and the range is −3 ≤ y ≤ 6.
Replacing x with −x reflects the graph in the y-axis, and subtracting 2 from the whole expression moves every point 2 units down. Because the vertical shift acts on the output, the order of these two operations does not change the final graph. Hence the combined transformation is a y-axis reflection followed by a downward shift of 2.
Completing the square gives f(x) = (x + 1)² + 4, so on x ≥ −1 the function is increasing with minimum value 4 at x = −1. Solving y = (x + 1)² + 4 for x gives x + 1 = ±√(y − 4), and the restriction x ≥ −1 forces the positive root. Swapping variables yields f⁻¹(x) = −1 + √(x − 4) with domain x ≥ 4.
The graph has its vertex at x = −3, where h(−3) = −2, and this vertex lies inside the interval −5 ≤ x ≤ 0, so −2 is the minimum value. Checking the endpoints gives h(−5) = 0 and h(0) = 1, so the largest value is 3, attained at x = −5. Hence the range is −2 ≤ h(x) ≤ 3.
The angle π/6 radians equals 30 degrees. In a 30-60-90 triangle the side opposite 30 degrees is half the hypotenuse, so sin(π/6)=1/2. This is one of the standard exact values worth memorising.
The tangent function has natural period π, and replacing x with kx compresses the period to π divided by |k|. Here k=3, so the period is π/3. Sine and cosine would instead have natural period 2π.
This is a complementary-angle identity: the sine of an angle equals the cosine of its complement. Since π/2 − θ and θ are complementary, sin(π/2 − θ) = cos θ. The identity holds for all real θ.
The angle 5π/6 lies in the second quadrant, where cosine is negative. Its reference angle is π/6, and cos(π/6)=√3/2, so cos(5π/6)=−√3/2.
Sine equals 1/2 at the first-quadrant angle π/6 and the second-quadrant angle π − π/6 = 5π/6. Because sine has period 2π, each base solution repeats with +2kπ. Both families together give the full solution set.
For y = a cos(bx) + c the amplitude is |a|, the distance from the midline to a peak. Here |−4| = 4. The negative sign reflects the graph but does not change the amplitude.
The angle 2π/3 is in the second quadrant where sine is positive. Its reference angle is π/3, and sin(π/3)=√3/2, so sin(2π/3)=√3/2.
As x approaches π/2 from below, tan x grows without bound, and from above it decreases without bound. Between consecutive asymptotes the tangent takes every real value, so its range is all real numbers.
Using the Pythagorean identity sin²θ + cos²θ = 1 gives 1 − cos²θ = sin²θ. Dividing by sin θ yields sin θ, valid wherever sin θ is nonzero.
The double-angle identity has three equivalent forms: cos 2θ = cos²θ − sin²θ = 2cos²θ − 1 = 1 − 2sin²θ. Substituting sin²θ = 1 − cos²θ into the first form gives the third.
Cosine is zero at π/2 and again at 3π/2, which differ by π. Since cosine has period 2π but the zeros repeat every π, the general solution is x = π/2 + kπ for integer k.
Both terms have the form with coefficient 4, so each has period 2π/4 = π/2. The period of a sum is the least common multiple of the component periods, which here is π/2 itself.
Writing 3θ = 2θ + θ and applying the sum and double-angle identities gives sin 3θ = sin θ(3 − 4sin²θ) = 3 sin θ − 4 sin³θ. This is the triple-angle formula for sine.
Factor as (2 sin x + 1)(sin x − 1) = 0, giving sin x = 1 or sin x = −1/2. The first gives x = π/2; the second gives x = 7π/6 and 11π/6 in the given interval.
The angle π + θ lies in the third quadrant for small positive θ, where sine is negative. Using sin(A + B) = sin A cos B + cos A sin B with A = π gives 0 − sin θ = −sin θ.
The equation is sin x = 2/3, and since 0 < 2/3 < 1 there are two solutions in one full period: one in the first quadrant and one in the second quadrant. Thus there are exactly two solutions.
For tan(kx) the period is π/|k|. Here k = 1/2, so the period is π ÷ (1/2) = 2π. Halving the argument stretches the graph horizontally.
For acute θ, cos θ = √(1 − 9/25) = 4/5. Then sin 2θ = 2 sin θ cos θ = 2 × (3/5) × (4/5) = 24/25. The value is less than 1 as required.
Expand: sin²x + 2 sin x cos x + cos²x. Since sin²x + cos²x = 1 and 2 sin x cos x = sin 2x, the expression equals 1 + sin 2x.
Write the argument as 3(x − π/3). A shift of π/3 to the right inside the bracket produces the −π term after multiplying by 3. Hence the phase shift is π/3 to the right.
The angle 7π/6 lies in the third quadrant, where tangent is positive. Its reference angle is π/6 and tan(π/6) = 1/√3, so tan(7π/6) = 1/√3.