AP Calculus AB Exam Question

Let f be the function defined by [f(x) = \frac{1}{x^{2}+1}].

a) Determine the average value of f(x) on the interval [-3,5].

b) Determine if there exists a value c in the interval [-3,5] such that f(c) = the average value found in part (a). Justify your answer.

Show all necessary work for full credit.

Answer and Explanation

a) To find the average value of a function on an interval, we need to evaluate the following definite integral:

Here, the interval is [-3,5], and the function is $f(x) = \frac{1}{x^{2}+1}$. Substituting these values, we have:

Integrating the function, we get: [\int_{-3}^{5} \frac{1}{x^{2}+1} , dx = \left[ \arctan(x) \right]_{-3}^{5} = \arctan(5) - \arctan(-3)]

Evaluating the arctangent function at $x=\pm3$ and simplifying, we have: [\arctan(5) - \arctan(-3) = \frac{\pi}{2} - \left(-\frac{\pi}{2}\right) = \pi]

Finally, substituting the value of the definite integral back into the average value formula, we get: [\text{Avg}(f) = \frac{1}{5 - (-3)} \cdot \pi = \frac{\pi}{8}]

Therefore, the average value of the function $f(x) = \frac{1}{x^{2}+1}$ on the interval [-3,5] is $\frac{\pi}{8}$.

b) To determine if there exists a value $c$ in the interval [-3,5] such that $f(c)$ equals the average value found in part (a), we need to find the solution(s) to the equation $f(c) = \frac{\pi}{8}$.

Substituting the function $f(x) = \frac{1}{x^{2}+1}$ into the equation, we have: [\frac{1}{c^{2}+1} = \frac{\pi}{8}]

To solve for $c$, we can Cross Multiply: [8 = \pi(c^{2}+1)]

Expand the expression: [8 = \pi c^{2} + \pi]

Rearranging the terms: [\pi c^{2} = 8 - \pi]

And finally, dividing by $\pi$: [c^{2} = \frac{8 - \pi}{\pi}]

At this point, we can see that the right-hand side expression will always be positive, since $\pi$ is approximately $3.14$ and $8 - \pi$ is greater than zero.

Thus, we have multiple solutions for $c$. Taking the square root, we get: [c = \pm \sqrt{\frac{8 - \pi}{\pi}}]

Therefore, there exist values $c$ in the interval [-3,5] such that $f(c) = \frac{\pi}{8}$.