What is the tangent of f(x)=(4x+1)^2,\at x=0 ?

The tangent of f(x)=(4x+1)^2,\at x=0 is y=8x+1

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

tangent of f(x)=(4x+1)^2,\at x=0

tangent of

f (x) = (4 x + 1)^{2}, at x = 0

y = 8 x + 1

Solution steps

Find the tangent point:

(0, 1)

Find the slope of

f (x) = (4 x + 1)^{2} : \frac{df ( x )}{d x} = 8 (4 x + 1)

Compute the slope of

f (x) = (4 x + 1)^{2}

x = 0 : m = 8

Find the line with slope m=

8

and passing through

(0, 1) : y = 8 x + 1

y = 8 x + 1

(x + 1) \cdot ln (x)

\int \frac{1}{e ^{x}} cos (π) x d x

y^{^{'}} = 10 x y

\int \frac{1}{20} (x + 1)^{2} (x - 3) (x + 4) d x

y^{^{''}} + 4 y^{^{'}} - 4 y = 0

What is the tangent of f(x)=(4x+1)^2,\at x=0 ?
The tangent of f(x)=(4x+1)^2,\at x=0 is y=8x+1