What is the tangent of f(x)=(x^2+7)(5x+4),\at x=2 ?

The tangent of f(x)=(x^2+7)(5x+4),\at x=2 is y=111x-68

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+7)(5x+4),\at x=2

tangent of

f (x) = (x^{2} + 7) (5 x + 4), at x = 2

y = 111 x - 68

Solution steps

Find the tangent point:

(2, 154)

Find the slope of

f (x) = (x^{2} + 7) (5 x + 4) : \frac{df ( x )}{d x} = 15 x^{2} + 8 x + 35

Compute the slope of

f (x) = (x^{2} + 7) (5 x + 4)

x = 2 : m = 111

Find the line with slope m=

111

and passing through

(2, 154) : y = 111 x - 68

y = 111 x - 68

x \to 0 lim (9)

\int \frac{1}{cos ^{2} ( 3 x )} d x

\int e

y^{^{'}} = \frac{2 x ^{7} e ^{\frac{y}{x}} + x ^{5} y ^{2}}{x ^{6} y}

\frac{d y}{d t} = 1 y, y (1) = 3

What is the tangent of f(x)=(x^2+7)(5x+4),\at x=2 ?
The tangent of f(x)=(x^2+7)(5x+4),\at x=2 is y=111x-68