What is the tangent of f(x)=3x^3-x^2+1,(2,21) ?

The tangent of f(x)=3x^3-x^2+1,(2,21) is y=32x-43

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tangent of f(x)=3x^3-x^2+1,(2,21)

tangent of

f (x) = 3 x^{3} - x^{2} + 1, (2, 21)

y = 32 x - 43

Solution steps

Find the slope of

f (x) = 3 x^{3} - x^{2} + 1 : \frac{df ( x )}{d x} = 9 x^{2} - 2 x

Compute the slope of

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

(2, 21) : m = 32

Find the line with slope m=

32

and passing through

(2, 21) : y = 32 x - 43

y = 32 x - 43

\int_{- 2}^{1} \frac{1}{x ^{2}} d x

\int (2 x)^{2} d x

(- 2, - 6), (2, 2)

f (x) = 9 + 4 x^{2} - 2 x^{3}, at x = 1

y = ln (ln (x^{3}))

What is the tangent of f(x)=3x^3-x^2+1,(2,21) ?
The tangent of f(x)=3x^3-x^2+1,(2,21) is y=32x-43