Question Find the linear approximation of $\cos 8$ about $x_0 = \frac{1}{2}$ for the function $f(x) = \cos(2x)$ .

Studdy Solution

STEP 1

Assumptions
1. The function given is $f(x) = \cos(2x)$ .
2. We need to find the linear approximation of $\cos(8)$ about $x_0 = \frac{1}{2}$ .
3. A linear approximation can be found using the formula for the tangent line at $x_0$ , which is $f(x_0) + f'(x_0)(x - x_0)$ .
4. The derivative of $\cos(2x)$ with respect to $x$ is needed for the linear approximation.

STEP 2

First, we need to find the derivative of $f(x) = \cos(2x)$ with respect to $x$ .
$f'(x) = \frac{d}{dx} \cos(2x)$

STEP 3

Using the chain rule, the derivative of $\cos(2x)$ is $-2\sin(2x)$ .
$f'(x) = -2\sin(2x)$

STEP 4

Now, we need to evaluate $f(x)$ and $f'(x)$ at $x_0 = \frac{1}{2}$ .
$f\left(\frac{1}{2}\right) = \cos\left(2 \cdot \frac{1}{2}\right)$

STEP 5

Calculate $f\left(\frac{1}{2}\right)$ .
$f\left(\frac{1}{2}\right) = \cos(1)$

STEP 6

Evaluate $f'(x)$ at $x_0 = \frac{1}{2}$ .
$f'\left(\frac{1}{2}\right) = -2\sin\left(2 \cdot \frac{1}{2}\right)$

STEP 7

Calculate $f'\left(\frac{1}{2}\right)$ .
$f'\left(\frac{1}{2}\right) = -2\sin(1)$

STEP 8

Now we have both $f\left(\frac{1}{2}\right)$ and $f'\left(\frac{1}{2}\right)$ . We can use these to find the linear approximation of $f(x)$ about $x_0 = \frac{1}{2}$ .
$L(x) = f\left(\frac{1}{2}\right) + f'\left(\frac{1}{2}\right)(x - \frac{1}{2})$

STEP 9

Substitute the values of $f\left(\frac{1}{2}\right)$ and $f'\left(\frac{1}{2}\right)$ into the linear approximation formula.
$L(x) = \cos(1) - 2\sin(1)(x - \frac{1}{2})$

STEP 10

Finally, we need to evaluate the linear approximation $L(x)$ at $x = \frac{8}{2} = 4$ to approximate $\cos(8)$ .
$L(4) = \cos(1) - 2\sin(1)(4 - \frac{1}{2})$

STEP 11

Calculate $L(4)$ .
$L(4) = \cos(1) - 2\sin(1)\left(\frac{7}{2}\right)$

STEP 12

Simplify the expression.
$L(4) = \cos(1) - 7\sin(1)$
The linear approximation of $\cos(8)$ about $x_0 = \frac{1}{2}$ is $\cos(1) - 7\sin(1)$ .

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