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

The tangent of f(x)=x^3-3x+2,\at x=3 is y=24x-52

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

tangent of

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

y = 24 x - 52

Solution steps

Find the tangent point:

(3, 20)

Find the slope of

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

Compute the slope of

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

x = 3 : m = 24

Find the line with slope m=

24

and passing through

(3, 20) : y = 24 x - 52

y = 24 x - 52

n = 1 \sum \infty 7 (\frac{1}{10})^{n}

\frac{d}{d x} (sin (1 - x))

\frac{1}{( s + 5 ) ^{2}}

\frac{d}{d x} (\frac{2 x ^{5} + 3 x ^{4} - x ^{3} + 2}{x ^{2}})

\int_{0}^{3} 2 x - 1 d x

What is the tangent of f(x)=x^3-3x+2,\at x=3 ?
The tangent of f(x)=x^3-3x+2,\at x=3 is y=24x-52