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

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

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

tangent of

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

y = 2 x + 1

Solution steps

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}

(1, 3) : m = 2

Find the line with slope m=

2

and passing through

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

y = 2 x + 1

\int - (1 - ln (x)) d x

\int \frac{x}{3} d x

x \to \infty lim (2 x^{2} - 1)

y^{^{'}} = y (x y^{3} + 1)

\frac{2}{x ^{2}}

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