Instantaneous Average Rate 245A5C

1. **State the problem:** We are given a function $h$ with derivative $h'(x) = 4e^{\cos x}$ and asked to find the value of $x$ in $[-3,0]$ where the instantaneous rate of change of $h$ equals the average rate of change of $f$ over the same interval. 2. **Clarify the problem:** The instantaneous rate of change of $h$ at $x$ is $h'(x)$. The average rate of change of $f$ over $[-3,0]$ is $\frac{f(0)-f(-3)}{0 - (-3)} = \frac{f(0)-f(-3)}{3}$. 3. **Note:** The problem asks about $h$'s instantaneous rate of change and $f$'s average rate of change. Since $h$ and $f$ are different functions, we need more information about $f$ to proceed. However, since only $h'(x)$ is given, we assume the problem intends to find $x$ where $h'(x)$ equals the average rate of change of $h$ over $[-3,0]$. 4. **Find the average rate of change of $h$ over $[-3,0]$:** $$\text{Average rate} = \frac{h(0) - h(-3)}{0 - (-3)} = \frac{h(0) - h(-3)}{3}$$ 5. **Find $h(0)$ and $h(-3)$:** Since $h'(x) = 4e^{\cos x}$, $h(x)$ is an antiderivative: $$h(x) = \int 4e^{\cos x} \, dx + C$$ We cannot find a closed form easily, but we can express the difference as an integral: $$h(0) - h(-3) = \int_{-3}^0 h'(x) \, dx = \int_{-3}^0 4e^{\cos x} \, dx$$ 6. **Calculate the average rate of change:** $$\frac{h(0) - h(-3)}{3} = \frac{1}{3} \int_{-3}^0 4e^{\cos x} \, dx = \frac{4}{3} \int_{-3}^0 e^{\cos x} \, dx$$ 7. **Set instantaneous rate equal to average rate:** $$h'(x) = 4e^{\cos x} = \frac{4}{3} \int_{-3}^0 e^{\cos t} \, dt$$ Divide both sides by 4: $$e^{\cos x} = \frac{1}{3} \int_{-3}^0 e^{\cos t} \, dt$$ 8. **Solve for $x$:** Take natural logarithm on both sides: $$\cos x = \ln \left( \frac{1}{3} \int_{-3}^0 e^{\cos t} \, dt \right)$$ 9. **Find $x$ in $[-3,0]$:** $$x = \arccos \left( \ln \left( \frac{1}{3} \int_{-3}^0 e^{\cos t} \, dt \right) \right)$$ 10. **Summary:** The value of $x$ where the instantaneous rate of change of $h$ equals the average rate of change of $h$ over $[-3,0]$ is given by the above expression. Numerical approximation requires evaluating the integral $\int_{-3}^0 e^{\cos t} dt$ and then computing $x$ accordingly.