What is the tangent of f(x)=6x^2-x^3,\at x=1 ?

The tangent of f(x)=6x^2-x^3,\at x=1 is y=9x-4

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tangent of f(x)=6x^2-x^3,\at x=1

tangent of

f (x) = 6 x^{2} - x^{3}, at x = 1

y = 9 x - 4

Solution steps

Find the tangent point:

(1, 5)

Find the slope of

f (x) = 6 x^{2} - x^{3} : \frac{df ( x )}{d x} = 12 x - 3 x^{2}

Compute the slope of

f (x) = 6 x^{2} - x^{3}

x = 1 : m = 9

Find the line with slope m=

9

and passing through

(1, 5) : y = 9 x - 4

y = 9 x - 4

- sin (x) - 4

\int (t) (e^{2 t}) d t

f (x) = cosh (3 \cdot ln (\frac{x}{2}))

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

\int \frac{1}{( 7 x - 1 ) ^{2}} d x

What is the tangent of f(x)=6x^2-x^3,\at x=1 ?
The tangent of f(x)=6x^2-x^3,\at x=1 is y=9x-4