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I am looking to find the linearization of $f(x) = 8\cos^2(x)$ at the point where $x = \pi/4$

Now, I know that the general formula for finding linearization is $L(x) = f(x) + f'(x) (x-a)$. So, I have made the following steps to get my answer:

1) Use the above formula to get $L(x) = 8\cos^2(\pi/4)+16\cos(\pi/4)(-\sin(\pi/4))(\pi/4-a)$

2) Simplify to get $4-8(\pi/4-a)$

However, I am not sure if this is completely correct.

Does anyone know whether this is the correct way to find this answer, and if it is not, how to arrive at the true answer?

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    $\begingroup$ You are to use $a= \pi / 4$ in $L(x) = f(a) + f'(a)(x-a).$ What you wrote makes no sense at all $\endgroup$
    – Will Jagy
    Nov 8, 2015 at 20:30

2 Answers 2

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First find the derivative:

$f'(x)=16\cos(x)(-\sin(x))=-16\sin(x)\cos(x)$.

Then $f'(\frac{\pi}{4})=-16\cos(\frac{\pi}{4})\sin(\frac{\pi}{4})=-16\frac{\sqrt2}{2}\cdot\frac{\sqrt2}{2}=-16\cdot\frac{2}{4}=-8$.

Then find the tangent: $L=f(\frac{\pi}{4})-8(x-\frac{\pi}{4})=4+2\pi-8x$

EDIT: I think your confusion is in the way you interpreted the formula for $L(x)$. It is poorly written and should be written more like this:

The tangent to $f(x)$ at $x=a$ is given by $L(x)=f(a)+f'(a)(x-a)$

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  • $\begingroup$ The answer, I believe, is $4-8x+2\pi$, but thank you for pointing out that my formula was wrong. $\endgroup$
    – Kelsey
    Nov 8, 2015 at 21:31
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here is a way to check your answer. $y = 8\cos^2 x = 4 + 4 \cos(2x).$ making a change of variable $x = \pi/4 + h,$ we have $y = 4 + 4\cos(\pi/2 + 2h)=4 - 4\sin(2h) = 4-8h+\cdots$ for $h = 0+ \cdots$

that is $$y = 8\cos^2 x = 4 - 8(x - \pi/4) + \cdots \text{ for } x = \pi/4 + \cdots $$

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