What is the tangent of f(x)=x^3-3x^2+1,\at x=2 ?

The tangent of f(x)=x^3-3x^2+1,\at x=2 is y=-3

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tangent of f(x)=x^3-3x^2+1,\at x=2

tangent of

f (x) = x^{3} - 3 x^{2} + 1, at x = 2

y = - 3

Solution steps

Find the tangent point:

(2, - 3)

Find the slope of

f (x) = x^{3} - 3 x^{2} + 1 : \frac{df ( x )}{d x} = 3 x^{2} - 6 x

Compute the slope of

f (x) = x^{3} - 3 x^{2} + 1

x = 2 : m = 0

Find the line with slope m=

0

and passing through

(2, - 3) : y = - 3

y = - 3

\int \frac{ln ( ln ( x ) )}{x} d x

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π r^{2} h

\int x^{3} - 4 x d x

\int sin^{9} (x) cos^{5} (x) d x

What is the tangent of f(x)=x^3-3x^2+1,\at x=2 ?
The tangent of f(x)=x^3-3x^2+1,\at x=2 is y=-3