What is the tangent of f(x)=7x^2-x^3,\at x=1 ?

The tangent of f(x)=7x^2-x^3,\at x=1 is y=11x-5

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tangent of f(x)=7x^2-x^3,\at x=1

tangent of

f (x) = 7 x^{2} - x^{3}, at x = 1

y = 11 x - 5

Solution steps

Find the tangent point:

(1, 6)

Find the slope of

f (x) = 7 x^{2} - x^{3} : \frac{df ( x )}{d x} = 14 x - 3 x^{2}

Compute the slope of

f (x) = 7 x^{2} - x^{3}

x = 1 : m = 11

Find the line with slope m=

11

and passing through

(1, 6) : y = 11 x - 5

y = 11 x - 5

\int_{0}^{1} ln (4 + x^{2}) d x

\int - u cos (2 u) d u

\int \frac{1}{[ ( x ^{2} + 9 ) ^{2} ]} d x

\frac{d}{d x} (\frac{7 ^{x}}{x ^{3}})

\int \frac{e ^{2 x} - 1}{e ^{2 x} + 3} d x

What is the tangent of f(x)=7x^2-x^3,\at x=1 ?
The tangent of f(x)=7x^2-x^3,\at x=1 is y=11x-5