Trig Derivative Problem

**scherz0** · January 26th 2009, 06:41 AM

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 $f(x) = x^2 + \cos(kx)$ , where k is the real parameter. For what values of k does the given function have no inflection points?

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I found these:

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

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

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

Therefore:

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

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

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

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

**red_dog** · January 26th 2009, 08:13 AM

The condition is $\frac{2}{k^2}>1$ , (because $\cos\alpha\in[-1,1], \ \forall\alpha\in\mathbf{R}$ )

Then $k^2<2\Rightarrow k\in(-\sqrt{2},\sqrt{2})$

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