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The square root requires ln x - 1 >= 0, that is ln x >= 1. Because the natural logarithm is increasing on its domain, this gives x >= e, and any such x automatically satisfies x > 0. Hence the domain is [e, infinity).
Writing y = e^(2x - 1) and taking logarithms gives ln y = 2x - 1, so x = (ln y + 1)/2. Interchanging x and y produces the inverse, and its domain is the range of f, namely x > 0.
As x varies, cos(4x) takes every value in [-1, 1]. Multiplying by -2 reverses the interval to [-2, 2], and adding 3 shifts it to [1, 5], which is therefore the range.
Both sides are non-negative, so squaring is reversible: (x - 3)^2 >= 4(x + 1)^2. Expanding and collecting gives 3x^2 + 14x - 5 <= 0, that is (3x - 1)(x + 5) <= 0, whose solution is -5 <= x <= 1/3.
For the monic cubic P(x) = x^3 + ax^2 + bx + 6, the product of the three roots equals -6. The known roots are 1 and -2, so 1 times -2 times r = -6, giving r = 3 and hence the factor x - 3.
The denominator factors as (x - 2)(x + 2) and shares no factor with the numerator (2x - 1)(x - 1), so the graph has vertical asymptotes x = 2 and x = -2. Since numerator and denominator have equal degree 2, the horizontal asymptote is the ratio of leading coefficients, y = 2.
Write 3x + 5 = A(x + 2) + B(x + 1). Substituting x = -1 gives 2 = A, and substituting x = -2 gives -1 = -B, so B = 1. The decomposition is therefore 2/(x + 1) + 1/(x + 2).
Substituting t = 3^x turns the equation into t^2 - 12t + 27 = 0, which factors as (t - 3)(t - 9) = 0. Since t must be positive, t = 3 or t = 9 give x = 1 or x = 2, whose sum is 3.
Continuity at x = 1 requires 1 + a + b = 3, so a + b = 2. Differentiability requires the one-sided derivatives to agree: 2x + a at x = 1 equals 2 + a, while the derivative of 2 ln x + 3 at x = 1 equals 2, so a = 0 and then b = 2.
Put t = e^x > 0, so the equation becomes t^2 - kt + k = 0, and we need exactly one positive root. A repeated root t = k/2 must be positive, forcing k = 4 (giving t = 2); alternatively the two roots have opposite signs exactly when their product k is negative, giving one positive root for every k < 0.
By the remainder theorem, the remainder on division by x - 2 equals P(2). Substituting x = 2 gives 2(8) - 5(4) + 4(2) - 7 = 16 - 20 + 8 - 7 = -3.
First compute f(x + 1) = (x + 1)^2 + 2(x + 1) = x^2 + 4x + 3. Subtracting f(x) = x^2 + 2x gives 2x + 3.
The term containing x^3 comes from choosing (2x)^3 from three factors and -1 from the other two. Its coefficient is C(5,3) times 2^3 times (-1)^2 = 10 times 8 = 80.
Combining the logarithms gives log_2((x - 1)(x - 3)) = 3, so (x - 1)(x - 3) = 8. This gives x^2 - 4x - 5 = 0, so x = 5 or x = -1; the logarithm domain x > 3 eliminates x = -1.
Write the decomposition as A/x + B/(x - 1) + C/(x - 1)^2. Multiplying by x(x - 1)^2 and equating coefficients gives A = 1, B = 0, and C = 2, so the decomposition is 1/x + 2/(x - 1)^2.
Because P has real coefficients, complex roots occur in conjugate pairs, so 1 - i is also a root. The product (x - (1 + i))(x - (1 - i)) equals x^2 - 2x + 2, and dividing P by this quadratic leaves the factor x - 2.
Let y = x/(x^2 + 1). Rearranging gives yx^2 - x + y = 0, which has a real solution x only when the discriminant 1 - 4y^2 is non-negative. Hence -1/2 <= y <= 1/2, and both endpoints are attained at x = 1 and x = -1.
Substituting x = 2 gives f(2) + 2f(1/2) = 2, while substituting x = 1/2 gives f(1/2) + 2f(2) = 1/2. Eliminating f(1/2) gives f(2) + 1 - 4f(2) = 2, so -3f(2) = 1 and f(2) = -1/3.
Rewrite the equation as x^2(x - 2)^2 = -a. The graph of g(x) = x^2(x - 2)^2 has local minima 0 at x = 0 and x = 2, a local maximum 1 at x = 1, and tends to infinity as x tends to plus or minus infinity. A horizontal line y = c cuts the graph four times exactly when 0 < c < 1, so 0 < -a < 1, giving -1 < a < 0.
Both x = 0 and x = 1 satisfy the equation. For x < 0, 2^x < 1 while 1 + x^2 >= 1, so no negative solution exists. For x > 1, the function h(x) = 2^x - (1 + x^2) is negative at x = 2 and x = 4 but positive at x = 5, and its single turning point after x = 1 gives exactly one further root between 4 and 5.
The logarithm is defined only when its argument is strictly positive, so 9 - x^2 > 0. This is equivalent to x^2 < 9, which gives the open interval (-3, 3); the endpoints are excluded because ln 0 is undefined.
Substitute f(x) into g: g(f(x)) = 2(x^2 + 1) - 3. Expanding gives 2x^2 + 2 - 3 = 2x^2 - 1.
The inverse value f^{-1}(65) is the real x satisfying x^3 + 1 = 65. Hence x^3 = 64, and the real cube root is x = 4.
Factor numerator and denominator: f(x)=((x-3)(x+3))/((x-3)(x+1)). The factor x-3 cancels, so the discontinuity at x=3 is removable; the limiting y-value is (3+3)/(3+1)=3/2.
Bring all terms to one side: (x-1)/(x+2)-2 = (-x-5)/(x+2) ≤ 0. Multiplying by -1 reverses the inequality to (x+5)/(x+2) ≥ 0. A sign chart with critical points -5 and -2 gives x ≤ -5 or x > -2.
Substituting 1-x for x gives f(1-x)+2f(x)=3(1-x). Solving these two equations simultaneously gives f(x)=2-3x, so f(2)=2-6=-4.
Since the coefficients are real, 1+2i is also a root. The sum of the three roots is 3, so the third root is 1; multiplying the roots gives -b=5, so b=-5, and the pairwise sum gives a=7. Therefore a+b=2.
For real x, sin x takes all values in [-1,1], so 2+sin x lies in [1,3]. Because ln is increasing, the range is [ln 1, ln 3]=[0, ln 3].
The derivative 3x^2-3 is zero at x=±1, giving the two turning points. Exactly two distinct real roots occur when one turning point lies on the x-axis: f(-1)=2+k=0 or f(1)=-2+k=0, so k=-2 or k=2.
The inflection condition f''(1)=0 gives 6+2a=0, so a=-3. The local-maximum condition f'(-1)=0 gives 3-2a+b=0, so b=-9; also c=2. Hence f(2)=8-12-18+2=-20.
Completing the square gives x^2 − 4x + 7 = (x − 2)^2 + 3. Since a square is never negative, the expression is at least 3, attained at x = 2. Therefore the minimum value is 3.
A quadratic has two distinct real solutions when its discriminant is positive. Here the discriminant is (k − 2)^2 − 36, so the condition is |k − 2| > 6. This gives k < −4 or k > 8.
The denominator x + 2 is zero at x = −2, giving a vertical asymptote there. The numerator and denominator have the same degree, so the horizontal asymptote is the ratio of leading coefficients, y = 3/1 = 3.
Let t = 3^x, so the equation becomes t^2 − 12t + 27 = 0, which factors as (t − 3)(t − 9) = 0. Thus 3^x = 3 or 3^x = 9, giving x = 1 or x = 2. The sum of the solutions is 3.
Since x − 1 is a factor, P(1) = 0, so 1 + a + b − 6 = 0 and a + b = 5. Since x + 2 is a factor, P(−2) = 0, which gives −8 + 4a − 2b − 6 = 0, that is 2a − b = 7. Solving the two equations gives a = 4 and b = 1.
Write f(2x − 6) as f(2(x − 3)). Replacing x by 2x compresses the graph horizontally by a factor of 1/2, and then replacing x by x − 3 translates it 3 units to the right. Doing the translation first would give f(2x − 3), which is a different graph.
Applying the function in turn: f(3) = 3/2, then f(3/2) = (3/2)/(1/2) = 3, then f(3) = 3/2 again. The function has period 2 under iteration, so three applications return the value to 3/2.
The critical points are x = 2 and x = −3, so solve on the three intervals. For x < −3 the equation becomes −2x − 1 = 7, giving x = −4; for x > 2 it becomes 2x + 1 = 7, giving x = 3; on −3 ≤ x ≤ 2 it becomes 5 = 7, which has no solution. The sum of the solutions is −4 + 3 = −1.
Let t = x^2 ≥ 0, so t^2 − 4t + k = 0. Exactly three distinct real x-values occur only when t = 0 is a root, contributing x = 0, while the other root for t is positive, contributing x = ±√t. Substituting t = 0 gives k = 0, and the equation then has t = 0 and t = 4, giving x = 0, 2 and −2.
Replacing x by 1/x in the given relation gives f(1/x) − 2f(x) = 1/x^2. Eliminating f(1/x) between the two equations leaves −3f(x) = x^2 + 2/x^2, so f(x) = −(x^2 + 2/x^2)/3. Substituting x = 2 gives f(2) = −(4 + 1/2)/3 = −3/2.
The radicand must be non-negative and the denominator cannot be zero. The critical values are x=−3 and x=1; testing intervals shows the quotient is positive for x<−3 and x>1, with x=1 included and x=−3 excluded.
Set y=(2x+1)/(x−3) and solve for x: y(x−3)=2x+1, so x(y−2)=3y+1 and x=(3y+1)/(y−2). Replacing y by x gives f⁻¹(x)=(3x+1)/(x−2).
By the remainder theorem, the remainder on division by x−2 is P(2). Substituting gives 8−16+10−7=−5.
Squaring both sides gives (x−1)²<4(x+1)², which simplifies to 3x²+10x+3>0. The roots are x=−3 and x=−1/3, and the quadratic is positive outside these roots.
Dividing gives x²+1=(x−1)(x+1)+2, so y=x+1+2/(x−1). As |x| becomes large, the fractional term tends to zero, leaving the oblique asymptote y=x+1.
Since x²+2≥2, the reciprocal satisfies 0<f(x)≤1/2. The value 1/2 occurs at x=0 and is included, while 0 is approached as |x|→∞ but never attained.
A double root at x=1 requires P(1)=0 and P′(1)=0. These give a+b+4=0 and 3a+2b+5=0, whose solution is a=3 and b=−7.
The domain requires x>1. Combining the logarithms gives (x+3)(x−1)=8, so x²+2x−11=0 and x=−1±2√3. Only −1+2√3 is greater than 1.
For x<0 the function h(x)=2^x−x² is strictly increasing from −∞ to 1, giving one negative solution. For x>0, the known solutions are x=2 and x=4, and derivative analysis shows there are no others, so there are three real solutions in total.
The cubic f(x)=0 has three real roots: one less than −1 and two between −1 and 3. For a level below −1 the graph meets it once, while for a level between −1 and 3 it meets it three times. Hence the total is 1+3+3=7.