Question

Here is a graph of the function y, both continuous and differentiable. For approximately what values of x between -5 and 5 is the second derivative zero?

Answer 1

From the graph you can determine the first derivative by looking at the turning points. These are where the first derivative is zero.

$y' = x(x-2)(x+2)(x-5)(x+5) \\ \\ y' = x^5 -29x^3 +100x$

Take derivative to get 2nd derivative and set equal to zero:
$y'' = 5x^4 -87x^2 +100 = 0$

Solve for x:
$x^2 = \frac{87 \pm \sqrt{87^2 -2000}}{10} \\ \\ x = \pm 1.11, \pm 4.02$