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}

EN : T i tl e G e n er a l Eq u a t i o n Sl o p e A t P o in t 2 Eq : m = 5

Find the line with slope m=

5

and passing through

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

y = 5 x - 2

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