What is the tangent of f(x)=x^3-2x+2,(3,23) ?

The tangent of f(x)=x^3-2x+2,(3,23) is y=25x-52

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tangent of f(x)=x^3-2x+2,(3,23)

tangent of

f (x) = x^{3} - 2 x + 2, (3, 23)

y = 25 x - 52

Solution steps

Find the slope of

f (x) = x^{3} - 2 x + 2 : \frac{df ( x )}{d x} = 3 x^{2} - 2

Compute the slope of

f (x) = x^{3} - 2 x + 2

(3, 23) : m = 25

Find the line with slope m=

25

and passing through

(3, 23) : y = 25 x - 52

y = 25 x - 52

\int_{0}^{x^{3}} (t + 1) (t - 2) d t

\frac{d}{d x} (\frac{x ^{4} + 2 x - 1}{5 x ^{2} - 2 x + 1})

\int \frac{x ^{2}}{x ^{2} + 4} d x

\frac{\partial}{\partial x} (x y^{3} - x)

\int \frac{1}{324 - x ^{2}} d x

What is the tangent of f(x)=x^3-2x+2,(3,23) ?
The tangent of f(x)=x^3-2x+2,(3,23) is y=25x-52