What is the tangent of f(x)=sqrt(x^2-x+13),\at x=4 ?

The tangent of f(x)=sqrt(x^2-x+13),\at x=4 is y= 7/10 x+11/5

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

tangent of f(x)=sqrt(x^2-x+13),\at x=4

tangent of

f (x) = x^{2} - x + 13, at x = 4

y = \frac{7}{10} x + \frac{11}{5}

Solution steps

Find the tangent point:

(4, 5)

Find the slope of

f (x) = x^{2} - x + 13 : \frac{df ( x )}{d x} = \frac{2 x - 1}{2 x ^{2} - x + 13}

Compute the slope of

f (x) = x^{2} - x + 13

x = 4 : m = \frac{7}{10}

Find the line with slope m=

\frac{7}{10}

and passing through

(4, 5) : y = \frac{7}{10} x + \frac{11}{5}

y = \frac{7}{10} x + \frac{11}{5}

f (x) = - \frac{4}{3} x^{\frac{2}{3}}

(y^{2} + 1) d x = y \cdot sec^{2} (x) d y

f (x) = \frac{x ^{2}}{3 + x}

y^{^{'}} + \frac{1}{x} y = 8 x^{2}

\frac{d}{d x} (arctan (x + 1))

What is the tangent of f(x)=sqrt(x^2-x+13),\at x=4 ?
The tangent of f(x)=sqrt(x^2-x+13),\at x=4 is y= 7/10 x+11/5