What is the tangent of f(x)=16sqrt(x),\at x=16 ?

The tangent of f(x)=16sqrt(x),\at x=16 is y=2x+32

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

tangent of f(x)=16sqrt(x),\at x=16

tangent of

f (x) = 16 x, at x = 16

y = 2 x + 32

Solution steps

Find the tangent point:

(16, 64)

Find the slope of

f (x) = 16 x : \frac{df ( x )}{d x} = \frac{8}{x}

Compute the slope of

f (x) = 16 x

x = 16 : m = 2

Find the line with slope m=

2

and passing through

(16, 64) : y = 2 x + 32

y = 2 x + 32

(2 x^{- 2} + 3)^{- 3}

\int \frac{1}{1 0 ^{5} x - x ^{2}} d x

\int 4 x^{\frac{1}{3}} + 2 x^{- \frac{1}{3}} d x

y^{^{'}} = y (x y^{6} + 3)

2 x^{^{''}} + 40 x = 0

What is the tangent of f(x)=16sqrt(x),\at x=16 ?
The tangent of f(x)=16sqrt(x),\at x=16 is y=2x+32