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Let h(x)=x^(5)-2x^(4)+10x^(2)-10 x". " Below is Omar's attempt to write a formal justification for the fact that there exists a value c where -2 < c < 2" such that "h^(')(c)=6". " Is Omar's justification complete? If not, why? Omar's justification: {:[h(-2)=-4" and "],[h(2)=20", so "],[(h(2)-h(-2))/(2-(-2))=6.]:} So, according to the mean value theorem, there exists a value c somewhere in the interval -2 < c < 2 such that h^(')(c)=6. Choose 1 answer: (A) Yes, Omar's justification is complete. (B) No, Omar didn't establish that the average rate of change of h over [-2,2] is equal to 6 . (C) No, Omar didn't establish that h is differentiable.

Let\newlineh(x)=x5−2x4+10x2−10x.  h(x)=x^{5}-2 x^{4}+10 x^{2}-10 x \text {. } \newlineBelow is Omar's attempt to write a formal justification for the fact that there exists a value c c where\newline\[ -2

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Q. Let\newlineh(x)=x5−2x4+10x2−10x.  h(x)=x^{5}-2 x^{4}+10 x^{2}-10 x \text {. } \newlineBelow is Omar's attempt to write a formal justification for the fact that there exists a value c c where\newline−2<c<2 such that h′(c)=6.  -2<c<2 \text { such that } h^{\prime}(c)=6 \text {. } \newlineIs Omar's justification complete? If not, why?\newlineOmar's justification:\newlineh(−2)=−4 and h(2)=20, so h(2)−h(−2)2−(−2)=6. \begin{array}{l} h(-2)=-4 \text { and } \\ h(2)=20 \text {, so } \\ \frac{h(2)-h(-2)}{2-(-2)}=6 . \end{array} \newlineSo, according to the mean value theorem, there exists a value c c somewhere in the interval −2<c<2 -2<c<2 such that h′(c)=6 h^{\prime}(c)=6 .\newlineChoose 11 answer:\newline(A) Yes, Omar's justification is complete.\newline(B) No, Omar didn't establish that the average rate of change of h h over [−2,2] [-2,2] is equal to 66 .\newline(C) No, Omar didn't establish that h h is differentiable.
  1. Verify Continuity: Omar's justification relies on the Mean Value Theorem (MVT), which states that if a function is continuous on the closed interval [a,b][a, b] and differentiable on the open interval (a,b)(a, b), then there exists at least one cc in (a,b)(a, b) such that the instantaneous rate of change at cc (the derivative h′(c)h'(c)) is equal to the average rate of change over [a,b][a, b]. We need to check if Omar's function h(x)h(x) meets these conditions and if his calculation of the average rate of change is correct.
  2. Confirm Differentiability: First, let's verify that h(x)h(x) is continuous on the closed interval [−2,2][-2, 2]. Since h(x)h(x) is a polynomial function, it is continuous everywhere, including on the interval [−2,2][-2, 2].
  3. Calculate Average Rate: Next, we need to confirm that h(x)h(x) is differentiable on the open interval (−2,2)(-2, 2). Again, because h(x)h(x) is a polynomial, it is differentiable everywhere, so it is differentiable on (−2,2)(-2, 2).
  4. Calculate h(−2)h(-2): Now, let's calculate the average rate of change of h(x)h(x) over the interval [−2,2][-2, 2]. This is done by evaluating the function at the endpoints and using the formula for the average rate of change: (h(b)−h(a))/(b−a)(h(b) - h(a)) / (b - a). Omar has already provided these values, but we should verify them.
  5. Calculate h(2)h(2): Calculate h(−2)h(-2):
    h(−2)=(−2)5−2(−2)4+10(−2)2−10(−2)h(-2) = (-2)^5 - 2(-2)^4 + 10(-2)^2 - 10(-2)
    h(−2)=−32−2(16)+10(4)+20h(-2) = -32 - 2(16) + 10(4) + 20
    h(−2)=−32−32+40+20h(-2) = -32 - 32 + 40 + 20
    h(−2)=−64+60h(-2) = -64 + 60
    h(−2)=−4h(-2) = -4
    Omar's calculation of h(−2)h(-2) is correct.
  6. Calculate Average Rate: Calculate h(2)h(2):h(2)=(2)5−2(2)4+10(2)2−10(2)h(2) = (2)^5 - 2(2)^4 + 10(2)^2 - 10(2)h(2)=32−2(16)+10(4)−20h(2) = 32 - 2(16) + 10(4) - 20h(2)=32−32+40−20h(2) = 32 - 32 + 40 - 20h(2)=0+40−20h(2) = 0 + 40 - 20h(2)=20h(2) = 20Omar's calculation of h(2)h(2) is correct.
  7. Apply Mean Value Theorem: Now, calculate the average rate of change: \newline(h(2) - h(-2)) / (2 - (-2)) = (20 - (-4)) / (2 - (-2))\(\newline= (20 + 4) / (2 + 2)\newline= 24 / 4\newline= 6\)\newlineOmar's calculation of the average rate of change is correct.
  8. Apply Mean Value Theorem: Now, calculate the average rate of change:\newline(h(2) - h(-2)) / (2 - (-2)) = (20 - (-4)) / (2 - (-2))\(\newline= (20 + 4) / (2 + 2)\newline= 24 / 4\newline= 6\newlineOmar's calculation of the average rate of change is correct.Since \$h(x)\) is continuous on \([-2, 2]\) and differentiable on \((-2, 2)\), and the average rate of change over the interval is \(6\), the Mean Value Theorem guarantees the existence of at least one \(c\) in the interval \((-2, 2)\) such that \(h'(c) = 6\). Omar's justification appears to be complete.

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