Answer
(a) By the Intermediate Value Theorem, there is a number $c$ in the interval $(0, 100)$ such that $f(c) = 0$
(b) $~~x = 70.347~~$ is a root of the equation.
Work Step by Step
(a) Let $f(x) = 100~e^{-x/100} - 0.01x^2$
When $x = 0$:
$f(0) = 100~e^{-0/100} - 0.01(0)^2 = 100$
When $x = 100$:
$f(100) = 100~e^{-100/100} - 0.01(100)^2 = -63.2$
The function $f(x)$ is continuous for all real numbers. $f(0) \gt 0$ and $f(100) \lt 0$. By the Intermediate Value Theorem, there is a number $c$ in the interval $(0, 100)$ such that $f(c) = 0$
(b) Let $f(x) = 100~e^{-x/100} - 0.01x^2$
We can use the zoom function on a graphing calculator to see that $f(x) = 0$ when $x \approx 70.347$
