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Let h(x)=x^(3)-6x^(2)-10 x and let c be the number that satisfies the Mean Value Theorem for h on the interval [-4,5]. What is c ? Choose 1 answer: (A) -2 (B) -1 (C) 1 (D) 3

Let h(x)=x3−6x2−10x h(x)=x^{3}-6 x^{2}-10 x and let c c be the number that satisfies the Mean Value Theorem for h h on the interval [−4,5] [-4,5] .\newlineWhat is c c ?\newlineChoose 11 answer:\newline(A) −2-2\newline(B) −1-1\newline(C) 11\newline(D) 33

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Full solution

Q. Let h(x)=x3−6x2−10x h(x)=x^{3}-6 x^{2}-10 x and let c c be the number that satisfies the Mean Value Theorem for h h on the interval [−4,5] [-4,5] .\newlineWhat is c c ?\newlineChoose 11 answer:\newline(A) −2-2\newline(B) −1-1\newline(C) 11\newline(D) 33
  1. Find h′(x)h'(x): To use the Mean Value Theorem, we need to find h′(x)h'(x), which is the derivative of h(x)h(x).
    h(x)=x3−6x2−10xh(x) = x^3 - 6x^2 - 10x
    h′(x)=3x2−12x−10h'(x) = 3x^2 - 12x - 10
  2. Calculate h(x)h(x) at endpoints: Now we need to find the values of h(x)h(x) at the endpoints of the interval [−4,5][-4,5].
    h(−4)=(−4)3−6(−4)2−10(−4)=−64−96+40=−120h(-4) = (-4)^3 - 6(-4)^2 - 10(-4) = -64 - 96 + 40 = -120
    h(5)=(5)3−6(5)2−10(5)=125−150−50=−75h(5) = (5)^3 - 6(5)^2 - 10(5) = 125 - 150 - 50 = -75
  3. Apply Mean Value Theorem: According to the Mean Value Theorem, there exists a cc in the interval (−4,5)(-4,5) such that h′(c)h'(c) is equal to the average rate of change of h(x)h(x) over [−4,5][-4,5].\newlineThe average rate of change is (h(5)−h(−4))/(5−(−4))(h(5) - h(-4)) / (5 - (-4)).\newline(−75−(−120))/(5−(−4))=45/9=5(−75 - (-120)) / (5 - (-4)) = 45 / 9 = 5
  4. Set up equation for cc: We set h′(c)h'(c) equal to the average rate of change we found: 3c2−12c−10=53c^2 - 12c - 10 = 5
  5. Factor quadratic equation: Now we solve for cc:3c2−12c−15=03c^2 - 12c - 15 = 0
  6. Factor quadratic equation: Now we solve for cc:3c2−12c−15=03c^2 - 12c - 15 = 0We can factor this quadratic equation or use the quadratic formula to find cc. Let's try to factor it first.(3c+3)(c−5)=0(3c + 3)(c - 5) = 0

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