What is the tangent of f(x)=x^2-1/x ,(2,3.5) ?

The tangent of f(x)=x^2-1/x ,(2,3.5) is y= 17/4 x-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)=x^2-1/x ,(2,3.5)

tangent of

f (x) = x^{2} - \frac{1}{x}, (2, 3.5)

y = \frac{17}{4} x - 5

Solution steps

Find the slope of

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

Compute the slope of

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

(2, 3.5) : m = \frac{17}{4}

Find the line with slope m=

\frac{17}{4}

and passing through

(2, 3.5) : y = \frac{17}{4} x - 5

y = \frac{17}{4} x - 5

n = 1 \sum \infty \frac{( ( - 7 ) ^{n - 1} )}{8 ^{n}}

y^{^{''}} - 14 y^{^{'}} + 49 y = t^{- 2} e^{7 t}

x \to - \infty lim (\frac{7}{x} - 4)

\int \frac{1}{w ^{3}} d w

12 x + 16 x^{\frac{3}{4}}

What is the tangent of f(x)=x^2-1/x ,(2,3.5) ?
The tangent of f(x)=x^2-1/x ,(2,3.5) is y= 17/4 x-5