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Let 
h(x)=3^(x)-x^(2).
Below is Uriah's attempt to write a formal justification for the fact that the equation 
h^(')(x)=8 has a solution where 
1 < x < 3.
Is Uriah's justification complete? If not, why?
Uriah's justification:
Polynomial and exponential functions are differentiable and continuous at all points in their domain, and 
[1,3] is within 
h 's domain. Also, 
h(1)=2 and 
h(3)=18, so

(h(3)-h(1))/(3-1)=8". "
So, according to the mean value theorem, 
h^(')(x)=8 must have a solution somewhere between 
x=1 and 
x=3.
Choose 1 answer:
(A) Yes, Uriah's justification is complete.
(B) No, Uriah didn't establish that the average rate of change of 
h over 
[1,3] is equal to 8 .
(C) No, Uriah didn't establish that 
h is differentiable.

Let h(x)=3xx2 h(x)=3^{x}-x^{2} .\newlineBelow is Uriah's attempt to write a formal justification for the fact that the equation h(x)=8 h^{\prime}(x)=8 has a solution where \( 1

Full solution

Q. Let h(x)=3xx2 h(x)=3^{x}-x^{2} .\newlineBelow is Uriah's attempt to write a formal justification for the fact that the equation h(x)=8 h^{\prime}(x)=8 has a solution where 1<x<3 1<x<3 .\newlineIs Uriah's justification complete? If not, why?\newlineUriah's justification:\newlinePolynomial and exponential functions are differentiable and continuous at all points in their domain, and [1,3] [1,3] is within h h 's domain. Also, h(1)=2 h(1)=2 and h(3)=18 h(3)=18 , so\newlineh(3)h(1)31=8 \frac{h(3)-h(1)}{3-1}=8 \text {. } \newlineSo, according to the mean value theorem, h(x)=8 h^{\prime}(x)=8 must have a solution somewhere between x=1 x=1 and x=3 x=3 .\newlineChoose 11 answer:\newline(A) Yes, Uriah's justification is complete.\newline(B) No, Uriah didn't establish that the average rate of change of h h over [1,3] [1,3] is equal to 88 .\newline(C) No, Uriah didn't establish that h h is differentiable.
  1. Polynomial and Exponential Functions: Uriah's justification begins by stating that polynomial and exponential functions are differentiable and continuous at all points in their domain, which is correct. Since h(x)h(x) is a combination of a polynomial and an exponential function, it is differentiable and continuous on its domain, which includes the interval [1,3][1,3].
  2. Calculation Verification: Uriah then states that h(1)=2h(1)=2 and h(3)=18h(3)=18. We need to check these calculations to ensure they are correct.\newlineh(1)=3112=31=2h(1) = 3^{1} - 1^{2} = 3 - 1 = 2\newlineh(3)=3332=279=18h(3) = 3^{3} - 3^{2} = 27 - 9 = 18\newlineThese calculations are correct.
  3. Average Rate of Change Calculation: Uriah uses these values to calculate the average rate of change of hh over the interval [1,3][1,3] as h(3)h(1)31=18231=162=8\frac{h(3)-h(1)}{3-1} = \frac{18-2}{3-1} = \frac{16}{2} = 8. This calculation is also correct.
  4. Mean Value Theorem Application: Uriah concludes that according to the mean value theorem, h(x)=8h^{\prime}(x)=8 must have a solution somewhere between x=1x=1 and x=3x=3. However, Uriah's justification is incomplete because he did not explicitly show that h(x)h^{\prime}(x) actually takes the value 88 at some point in the interval. The mean value theorem states that there exists some cc in (a,b)(a, b) such that f(c)f^{\prime}(c) is equal to the average rate of change of ff over [a,b][a, b]. Uriah has shown the average rate of change is 88, but he has not shown that x=1x=111 for some x=1x=122 in x=1x=133.

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