What is the tangent of f(x)=8x^2-x^3,\at x=1 ?

The tangent of f(x)=8x^2-x^3,\at x=1 is y=13x-6

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tangent of f(x)=8x^2-x^3,\at x=1

tangent of

f (x) = 8 x^{2} - x^{3}, at x = 1

y = 13 x - 6

Solution steps

Find the tangent point:

(1, 7)

Find the slope of

f (x) = 8 x^{2} - x^{3} : \frac{df ( x )}{d x} = 16 x - 3 x^{2}

Compute the slope of

f (x) = 8 x^{2} - x^{3}

x = 1 : m = 13

Find the line with slope m=

13

and passing through

(1, 7) : y = 13 x - 6

y = 13 x - 6

\int - x^{2} + 5 x - 4 d x

x \cdot \frac{d y}{d x} + y = \frac{1}{y ^{2}}

\int_{0}^{e} x^{2} ln (x) d x

\frac{d}{d x} (ln (x + e^{x}))

y^{^{'}} = \frac{( 1 - y ) ^{5}}{4}

What is the tangent of f(x)=8x^2-x^3,\at x=1 ?
The tangent of f(x)=8x^2-x^3,\at x=1 is y=13x-6