What is the tangent of f(x)=1x-32sqrt(x),\at x=4 ?

The tangent of f(x)=1x-32sqrt(x),\at x=4 is y=-7x-32

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tangent of f(x)=1x-32sqrt(x),\at x=4

tangent of

f (x) = 1 x - 32 x, at x = 4

y = - 7 x - 32

Solution steps

Find the tangent point:

(4, - 60)

Find the slope of

f (x) = 1 x - 32 x : \frac{df ( x )}{d x} = 1 - \frac{16}{x}

Compute the slope of

f (x) = 1 x - 32 x

x = 4 : m = - 7

Find the line with slope m=

- 7

and passing through

(4, - 60) : y = - 7 x - 32

y = - 7 x - 32

x \to \infty lim (x^{2} e^{x})

\frac{2}{s ( s ^{2} + 15 s + 50 )}

\int x ln (y + 1) d y

n = 0 \sum \infty (\frac{A}{A + 1})^{n}

What is the tangent of f(x)=1x-32sqrt(x),\at x=4 ?
The tangent of f(x)=1x-32sqrt(x),\at x=4 is y=-7x-32