What is the tangent of f(x)=x^2-sqrt(x),(1,0) ?

The tangent of f(x)=x^2-sqrt(x),(1,0) is y= 3/2 x-3/2

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-sqrt(x),(1,0)

tangent of

f (x) = x^{2} - x, (1, 0)

y = \frac{3}{2} x - \frac{3}{2}

Solution steps

Find the slope of

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

Compute the slope of

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

(1, 0) : m = \frac{3}{2}

Find the line with slope m=

\frac{3}{2}

and passing through

(1, 0) : y = \frac{3}{2} x - \frac{3}{2}

y = \frac{3}{2} x - \frac{3}{2}

x \to 9 lim (\frac{( x ^{2} - 4 x - 45 )}{x ^{2} - 81})

\int - 2 cos (t) sin^{2} (t) d t

\frac{d}{d x} (12 x^{2} + 4)

(- 1, - 2), (- 2, 3)

\int \frac{1}{sin ( x ) + 1} d x

What is the tangent of f(x)=x^2-sqrt(x),(1,0) ?
The tangent of f(x)=x^2-sqrt(x),(1,0) is y= 3/2 x-3/2