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

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

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

tangent of

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

y = 2 x + 1

Solution steps

Find the tangent point:

(1, 3)

Find the slope of

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

Compute the slope of

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

x = 1 : m = 2

Find the line with slope m=

2

and passing through

(1, 3) : y = 2 x + 1

y = 2 x + 1

x \to - 2 - lim (\frac{x ^{2} - 4}{x + 2})

\int x x^{2} - 2 d x

x \to 0 lim (\frac{π}{x})

f (x) = x x^{2} + 21

\int_{- e}^{0} e^{2 x} d x

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