What is the tangent of f(x)=x^2-4,\at x=5 ?

The tangent of f(x)=x^2-4,\at x=5 is y=10x-29

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

tangent of f(x)=x^2-4,\at x=5

tangent of

f (x) = x^{2} - 4, at x = 5

y = 10 x - 29

Solution steps

Find the tangent point:

(5, 21)

Find the slope of

f (x) = x^{2} - 4 : \frac{df ( x )}{d x} = 2 x

Compute the slope of

f (x) = x^{2} - 4

x = 5 : m = 10

Find the line with slope m=

10

and passing through

(5, 21) : y = 10 x - 29

y = 10 x - 29

n = 0 \sum \infty 3^{- 2 n}

\frac{d}{d x} (ln (3 x e^{- x}))

15 - x^{2}, - 6 x - 1, [- 4, - 1]

\int \frac{9}{( x + 1 ) ^{2}} d x

\int \frac{1 - 2 x}{( x + 1 ) ( x - 5 )} d x

What is the tangent of f(x)=x^2-4,\at x=5 ?
The tangent of f(x)=x^2-4,\at x=5 is y=10x-29