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

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

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tangent of f(x)=5x-x^2,\at x=1

tangent of

f (x) = 5 x - x^{2}, at x = 1

y = 3 x + 1

Solution steps

Find the tangent point:

(1, 4)

Find the slope of

f (x) = 5 x - x^{2} : \frac{df ( x )}{d x} = 5 - 2 x

Compute the slope of

f (x) = 5 x - x^{2}

x = 1 : m = 3

Find the line with slope m=

3

and passing through

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

y = 3 x + 1

y^{^{'}} = 2 (1 + x) (1 + y^{2}), y (0) = 0

\int (x^{2} + 3 x - 1) \cdot e^{2 x} d x

\frac{\partial}{\partial x} (5 x^{2 y})

\frac{d}{d x} (ln (csc (x) - cot (x)))

\frac{1}{x}

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