What is the tangent of f(x)=8sqrt(x),(16,32) ?

The tangent of f(x)=8sqrt(x),(16,32) is y=x+16

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

tangent of f(x)=8sqrt(x),(16,32)

tangent of

f (x) = 8 x, (16, 32)

y = x + 16

Solution steps

Find the slope of

f (x) = 8 x : \frac{df ( x )}{d x} = \frac{4}{x}

Compute the slope of

f (x) = 8 x

(16, 32) : m = 1

Find the line with slope m=

1

and passing through

(16, 32) : y = x + 16

y = x + 16

y = ln (3 x^{2} cot (sec^{2} (4 x)))

- 8 x^{2} - 2 x

t \to 1 lim (\frac{e ^{t} - e}{t - 1})

\frac{d}{d x} (x^{2} - 16 I n (x^{2}))

n = 2 \sum \infty \frac{5}{6 ^{n}}

What is the tangent of f(x)=8sqrt(x),(16,32) ?
The tangent of f(x)=8sqrt(x),(16,32) is y=x+16