What is the slope of the line normal to the tangent line of $f (x) = \frac{1}{x} + x^{3}$ $f(x) = 1/x+x^3$ at $x = 2$ $x= 2$ ?

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Answer 1 · 2016-09-17T02:51:27

$- \frac{4}{47}$ $-4/47$

The slope of the line tangent to the curve $y = f (x)$ $y=f(x)$ at $x = 2$ $x=2$ is given by $f'(2)$ .

First, find $f'(x)$ .

$f'(x) = frac{d}{dx}(1/x + x^3)$

$= -1/x^2 +3x^2$

Find $f'(2)$ by substituting $x=2$ .

$f'(2) = -1/2^2 + 3(2)^2$

$= 47/4$

The slope of the line normal to the tangent line, $m$ , is given by

$m = -1/(f'(2))$

$= - 4/47$

What is the slope of the line normal to the tangent line of f(x) = 1/x+x^3 f(x)=1x+x3f(x) = 1/x+x^3 at x= 2 x=2 x= 2 ?