What is the tangent of f(x)=x^3-5x+3,\at x=3 ?

The tangent of f(x)=x^3-5x+3,\at x=3 is y=22x-51

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tangent of f(x)=x^3-5x+3,\at x=3

tangent of

f (x) = x^{3} - 5 x + 3, at x = 3

y = 22 x - 51

Solution steps

Find the tangent point:

(3, 15)

Find the slope of

f (x) = x^{3} - 5 x + 3 : \frac{df ( x )}{d x} = 3 x^{2} - 5

Compute the slope of

f (x) = x^{3} - 5 x + 3

x = 3 : m = 22

Find the line with slope m=

22

and passing through

(3, 15) : y = 22 x - 51

y = 22 x - 51

f (x) = (- x^{2} - 4)^{2}

\frac{d}{d x} (3 x^{2} - \frac{1}{24} ln (x))

\int \frac{1}{x ^{2} + 4 x + 20} d x

x \to 1 lim (e^{x})

x \to 5 lim (x^{- 1})

What is the tangent of f(x)=x^3-5x+3,\at x=3 ?
The tangent of f(x)=x^3-5x+3,\at x=3 is y=22x-51