What is the tangent of f(x)=sqrt(x+24),(1,5) ?

The tangent of f(x)=sqrt(x+24),(1,5) is y= 1/10 x+49/10

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+24),(1,5)

tangent of

f (x) = x + 24, (1, 5)

y = \frac{1}{10} x + \frac{49}{10}

Solution steps

Find the slope of

f (x) = x + 24 : \frac{df ( x )}{d x} = \frac{1}{2 x + 24}

Compute the slope of

f (x) = x + 24

(1, 5) : m = \frac{1}{10}

Find the line with slope m=

\frac{1}{10}

and passing through

(1, 5) : y = \frac{1}{10} x + \frac{49}{10}

y = \frac{1}{10} x + \frac{49}{10}

\frac{\partial}{\partial y} (3 x^{2} + 3 y^{2})

\frac{\partial}{\partial y} (\frac{3}{x ^{2} + y ^{2}})

\frac{d}{d x} (3 x^{3} (x^{2} + 4)^{4})

\int 9 - A^{2} d A

x = y^{2} - 2 y, x - y - 4 = 0

What is the tangent of f(x)=sqrt(x+24),(1,5) ?
The tangent of f(x)=sqrt(x+24),(1,5) is y= 1/10 x+49/10