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

The tangent of f(x)=x^2+x,\at a=5 is y=(2a+1)x-a^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)=x^2+x,\at a=5

tangent of

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

y = (2 a + 1) x - a^{2}

Solution steps

Compute the tangent line to the general point

x = a

Find the tangent point:

(a, a^{2} + a)

Find the slope of

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

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 a + 1

Find the line with slope m=

2 a + 1

and passing through

(a, a^{2} + a) : y = (2 a + 1) x - a^{2}

y = (2 a + 1) x - a^{2}

\int (12 x + 14) (3 x^{2} + 7 x - 1)^{5} d x

n = 1 \sum \infty \frac{6}{n} - \frac{6}{n + 1}

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

y = \frac{1}{2} x^{2} + 9 x

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

What is the tangent of f(x)=x^2+x,\at a=5 ?
The tangent of f(x)=x^2+x,\at a=5 is y=(2a+1)x-a^2