What is the tangent of f(x)=-2x^4+5x^2-3,(1,0) ?

The tangent of f(x)=-2x^4+5x^2-3,(1,0) is y=2x-2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

tangent of f(x)=-2x^4+5x^2-3,(1,0)

tangent of

f (x) = - 2 x^{4} + 5 x^{2} - 3, (1, 0)

y = 2 x - 2

Solution steps

Find the slope of

f (x) = - 2 x^{4} + 5 x^{2} - 3 : \frac{df ( x )}{d x} = - 8 x^{3} + 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 = 2

Find the line with slope m=

2

and passing through

(1, 0) : y = 2 x - 2

y = 2 x - 2

\frac{d ^{2}}{d x ^{2}} (\frac{1}{x})

\frac{\partial}{\partial x} (\frac{y}{2 x + y})

\frac{\partial}{\partial x} (6 x y e^{x z})

x \to 0 lim (\frac{tan ( \frac{x}{3} )}{x})

x^{2} - 8 x

What is the tangent of f(x)=-2x^4+5x^2-3,(1,0) ?
The tangent of f(x)=-2x^4+5x^2-3,(1,0) is y=2x-2