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

The tangent of f(x)=4x^2-x^3,\at x=1 is y=5x-2

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

tangent of

f (x) = 4 x^{2} - x^{3}, at x = 1

y = 5 x - 2

Solution steps

Find the tangent point:

(1, 3)

Find the slope of

f (x) = 4 x^{2} - x^{3} : \frac{df ( x )}{d x} = 8 x - 3 x^{2}

Compute the slope of

f (x) = 4 x^{2} - x^{3}

x = 1 : m = 5

Find the line with slope m=

5

and passing through

(1, 3) : y = 5 x - 2

y = 5 x - 2

\frac{d y}{d x} + 2 y = 3

x \to 0 lim (\frac{1}{2} x^{2})

x \to 1 lim (3 x + 5)

3 csc (x) + 7 cos (x)

\int 7 (tan^{2} (x) + tan^{4} (x)) d x

What is the tangent of f(x)=4x^2-x^3,\at x=1 ?
The tangent of f(x)=4x^2-x^3,\at x=1 is y=5x-2