What is the tangent of f(x)=4x-x^2,\at x=0 ?

The tangent of f(x)=4x-x^2,\at x=0 is y=4x

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tangent of f(x)=4x-x^2,\at x=0

tangent of

f (x) = 4 x - x^{2}, at x = 0

y = 4 x

Solution steps

Find the tangent point:

(0, 0)

Find the slope of

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

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 = 4

Find the line with slope m=

4

and passing through

(0, 0) : y = 4 x

y = 4 x

\frac{d}{d x} (1 - cos (\frac{x}{2}))

\frac{d}{d x} (261 e^{0, 5 x} + 27.464)

\frac{d}{d x} (e^{x f (x)})

\frac{d}{d x} (e^{- 4 x} cos^{4} (3 x))

\int sec (t) (4 sec (t) + 3 tan (t)) d t

What is the tangent of f(x)=4x-x^2,\at x=0 ?
The tangent of f(x)=4x-x^2,\at x=0 is y=4x