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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?

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 class=

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wizard123
From the graph you can determine the first derivative by looking at the turning points. These are where the first derivative is zero.

[tex]y' = x(x-2)(x+2)(x-5)(x+5) \\ \\ y' = x^5 -29x^3 +100x[/tex]

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

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

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