What is the tangent of f(x)=sqrt(x^2+40),\at x=3 ?

The tangent of f(x)=sqrt(x^2+40),\at x=3 is y= 3/7 x+40/7

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

tangent of f(x)=sqrt(x^2+40),\at x=3

tangent of

f (x) = x^{2} + 40, at x = 3

y = \frac{3}{7} x + \frac{40}{7}

Solution steps

Find the tangent point:

(3, 7)

Find the slope of

f (x) = x^{2} + 40 : \frac{df ( x )}{d x} = \frac{x}{x ^{2} + 40}

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 = \frac{3}{7}

Find the line with slope m=

\frac{3}{7}

and passing through

(3, 7) : y = \frac{3}{7} x + \frac{40}{7}

y = \frac{3}{7} x + \frac{40}{7}

What is the tangent of f(x)=sqrt(x^2+40),\at x=3 ?
The tangent of f(x)=sqrt(x^2+40),\at x=3 is y= 3/7 x+40/7