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

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

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

tangent of

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

y = 4 x - 1

Solution steps

Find the tangent point:

(1, 3)

Find the slope of

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

Compute the slope of

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

x = 1 : m = 4

Find the line with slope m=

4

and passing through

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

y = 4 x - 1

\int ln (t) d t

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

11 x, x^{2} - 12

(- 1, - 1), (0, 0)

sin (x + \frac{π}{2})

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