Hello everyone,

Could anyone please offer hints or tips as to how to solve this problem? Some of my work is shown below.

Thank you.

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5. Consider the function $\displaystyle f(x) = x^2 + \cos(kx) $, wherekis the real parameter. For what values ofkdoes the given function have no inflection points?

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I found these:

$\displaystyle f'(x) = 2x - k\sin(kx) $

$\displaystyle f''(x) = 2 - k^2 \cos(kx)$

To determine the inflection points, I set $\displaystyle f''(x) = 0 $.

Therefore:

$\displaystyle k^2 \cos(kx) = 2 $

$\displaystyle cos(kx) = \frac{2}{k^2} $

$\displaystyle x = \frac{\cos^{-1} \frac{2}{k^2}}{k} $

However, I am not sure how to proceed from here.