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

The tangent of f(x)=-2x^3+x^2,\at x=1 is y=-4x+3

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

tangent of

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

y = - 4 x + 3

Solution steps

Find the tangent point:

(1, - 1)

Find the slope of

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

Compute the slope of

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

x = 1 : m = - 4

Find the line with slope m=

- 4

and passing through

(1, - 1) : y = - 4 x + 3

y = - 4 x + 3

y = 10 x, (10, 10)

\frac{d}{d x} (ln (5 - x^{2}))

\int_{2}^{4} \frac{1}{16 - t ^{2}} d t

x \to 2 lim (2 x - 2)

\int \frac{x ^{3}}{x ^{2} - 2 x + 4} d x

What is the tangent of f(x)=-2x^3+x^2,\at x=1 ?
The tangent of f(x)=-2x^3+x^2,\at x=1 is y=-4x+3