What is the tangent of f(x)=x^4-32x^2+256,\at (1,) ?

The tangent of f(x)=x^4-32x^2+256,\at (1,) is y=(4a_{0}^3-64a_{0})x-3a_{0}^4+32a_{0}^2+256

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^4-32x^2+256,\at (1,)

tangent of

f (x) = x^{4} - 32 x^{2} + 256, at (1,)

y = (4 a_{0}^{3} - 64 a_{0}) x - 3 a_{0}^{4} + 32 a_{0}^{2} + 256

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, a_{0}^{4} - 32 a_{0}^{2} + 256)

Find the slope of

f (x) = x^{4} - 32 x^{2} + 256 : \frac{df ( x )}{d x} = 4 x^{3} - 64 x

Compute the slope of

f (x) = x^{4} - 32 x^{2} + 256

x = a_{0} : m = 4 a_{0}^{3} - 64 a_{0}

Find the line with slope m=

4 a_{0}^{3} - 64 a_{0}

and passing through

(a_{0}, a_{0}^{4} - 32 a_{0}^{2} + 256) : y = (4 a_{0}^{3} - 64 a_{0}) x - 3 a_{0}^{4} + 32 a_{0}^{2} + 256

y = (4 a_{0}^{3} - 64 a_{0}) x - 3 a_{0}^{4} + 32 a_{0}^{2} + 256

\int cos (θ) d θ

\int \frac{n}{n + 1} d n

\int (3 x + \frac{10}{x ^{6}}) d x

f (x) = 3 x^{\frac{1}{2}} + x^{\frac{3}{2}}, at x = 9

\int 6 sin (2 t) d t

What is the tangent of f(x)=x^4-32x^2+256,\at (1,) ?
The tangent of f(x)=x^4-32x^2+256,\at (1,) is y=(4a_{0}^3-64a_{0})x-3a_{0}^4+32a_{0}^2+256