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

The tangent of f(x)=5x^2-x^3,\at x=1 is y=7x-3

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

tangent of

f (x) = 5 x^{2} - x^{3}, at x = 1

y = 7 x - 3

Solution steps

Find the tangent point:

(1, 4)

Find the slope of

f (x) = 5 x^{2} - x^{3} : \frac{df ( x )}{d x} = 10 x - 3 x^{2}

Compute the slope of

f (x) = 5 x^{2} - x^{3}

x = 1 : m = 7

Find the line with slope m=

7

and passing through

(1, 4) : y = 7 x - 3

y = 7 x - 3

2 sin (2 v) - 4 sin^{9} (2 v)

\frac{d}{d x} (x e^{- y})

\int_{- 2}^{- 1} \frac{2}{x ^{2}} d x

(4 x + 2 x^{- 1} y + 2 x^{- 1} y^{2}) d x + (1 + 2 y) d y = 0

f (x) = \frac{x}{( 1 + x ^{2} )}

What is the tangent of f(x)=5x^2-x^3,\at x=1 ?
The tangent of f(x)=5x^2-x^3,\at x=1 is y=7x-3