What is the tangent of f(x)=5x^2,\at x=0 ?

The tangent of f(x)=5x^2,\at x=0 is y=0

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

tangent of f(x)=5x^2,\at x=0

tangent of

f (x) = 5 x^{2}, at x = 0

y = 0

Solution steps

Find the tangent point:

(0, 0)

Find the slope of

f (x) = 5 x^{2} : \frac{df ( x )}{d x} = 10 x

EN : T i tl e G e n er a l Eq u a t i o n Sl o p e A t P o in t 2 Eq : m = 0

Find the line with slope m=

0

and passing through

(0, 0) : y = 0

y = 0

n = 1 \sum \infty (- 1)^{n + 1} \frac{n + 1}{2 n ^{2} + 1}

\int \frac{ln ( x ^{\frac{7}{2}} )}{x} d x

\frac{d}{d x} ((ln (x))^{p (x)})

\frac{d y}{d x} = \frac{y}{( x ^{2} + 1 )}

\frac{16}{s ( s + 2 ) ^{4}}

What is the tangent of f(x)=5x^2,\at x=0 ?
The tangent of f(x)=5x^2,\at x=0 is y=0