What is the tangent of f(x)=x^3-9x+5,\at x=3 ?

The tangent of f(x)=x^3-9x+5,\at x=3 is y=18x-49

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tangent of f(x)=x^3-9x+5,\at x=3

tangent of

f (x) = x^{3} - 9 x + 5, at x = 3

y = 18 x - 49

Solution steps

Find the tangent point:

(3, 5)

Find the slope of

f (x) = x^{3} - 9 x + 5 : \frac{df ( x )}{d x} = 3 x^{2} - 9

Compute the slope of

f (x) = x^{3} - 9 x + 5

x = 3 : m = 18

Find the line with slope m=

18

and passing through

(3, 5) : y = 18 x - 49

y = 18 x - 49

y = sec (x^{2} + 3 x)

\int 2 x (x^{2} - 7)^{104} d x

\frac{d}{d x} (15 ln (x) + 4 cos (x) + o)

\frac{d}{d x} ((4 x^{2} - 3 x) (2 x + 1))

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

What is the tangent of f(x)=x^3-9x+5,\at x=3 ?
The tangent of f(x)=x^3-9x+5,\at x=3 is y=18x-49