Find the Slope of the Tangent Line at x=-2 f(x)=x^4-2x^2+1 ; x=-2

$f (x) = x^{4} - 2 x^{2} + 1$ ; $x = - 2$

Step 1

Differentiate.

By the Sum Rule, the derivative of $x^{4} - 2 x^{2} + 1$ with respect to $x$ is $\frac{d}{d x} [x^{4}] + \frac{d}{d x} [- 2 x^{2}] + \frac{d}{d x} [1]$ .

$\frac{d}{d x} [x^{4}] + \frac{d}{d x} [- 2 x^{2}] + \frac{d}{d x} [1]$

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 4$ .

$4 x^{3} + \frac{d}{d x} [- 2 x^{2}] + \frac{d}{d x} [1]$

Step 2

Evaluate $\frac{d}{d x} [- 2 x^{2}]$ .

Since $- 2$ is constant with respect to $x$ , the derivative of $- 2 x^{2}$ with respect to $x$ is $- 2 \frac{d}{d x} [x^{2}]$ .

$4 x^{3} - 2 \frac{d}{d x} [x^{2}] + \frac{d}{d x} [1]$

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 2$ .

$4 x^{3} - 2 (2 x) + \frac{d}{d x} [1]$

Multiply $2$ by $- 2$ .

$4 x^{3} - 4 x + \frac{d}{d x} [1]$

Step 3

Differentiate using the Constant Rule.

Since $1$ is constant with respect to $x$ , the derivative of $1$ with respect to $x$ is $0$ .

$4 x^{3} - 4 x + 0$

Add $4 x^{3} - 4 x$ and $0$ .

$4 x^{3} - 4 x$

Step 4

Evaluate the derivative at $x = - 2$ .

$4 {(- 2)}^{3} - 4 \cdot - 2$

Step 5

Simplify.

Simplify each term.

Raise $- 2$ to the power of $3$ .

$4 \cdot - 8 - 4 \cdot - 2$

Multiply $4$ by $- 8$ .

$- 32 - 4 \cdot - 2$

Multiply $- 4$ by $- 2$ .

$- 32 + 8$

Add $- 32$ and $8$ .

$- 24$

Find the Slope of the Tangent Line at x=-2 f(x)=x^4-2x^2+1 ; x=-2

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