Q. Let h(x)=3x−x2.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, so3−1h(3)−h(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.
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) is a combination of a polynomial and an exponential function, it is differentiable and continuous on its domain, which includes the interval [1,3].
Calculation Verification: Uriah then states that h(1)=2 and h(3)=18. We need to check these calculations to ensure they are correct.h(1)=31−12=3−1=2h(3)=33−32=27−9=18These calculations are correct.
Average Rate of Change Calculation: Uriah uses these values to calculate the average rate of change of h over the interval [1,3] as 3−1h(3)−h(1)=3−118−2=216=8. This calculation is also correct.
Mean Value Theorem Application: Uriah concludes that according to the mean value theorem, h′(x)=8 must have a solution somewhere between x=1 and x=3. However, Uriah's justification is incomplete because he did not explicitly show that h′(x) actually takes the value 8 at some point in the interval. The mean value theorem states that there exists some c in (a,b) such that f′(c) is equal to the average rate of change of f over [a,b]. Uriah has shown the average rate of change is 8, but he has not shown that x=11 for some x=12 in x=13.
More problems from One-step inequalities: word problems