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

The tangent of f(x)=4x^2-x^3,(1,3) is y=5x-2

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

tangent of

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

y = 5 x - 2

Solution steps

Find the slope of

f (x) = 4 x^{2} - x^{3} : \frac{df ( x )}{d x} = 8 x - 3 x^{2}

Compute the slope of

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

(1, 3) : m = 5

Find the line with slope m=

5

and passing through

(1, 3) : y = 5 x - 2

y = 5 x - 2

y = - x^{2} - 2 x + 2, (- 3, - 1)

sin (x), 0

y^{^{'}} - \frac{9}{x} y = \frac{y ^{5}}{x ^{6}}

n = 1 \sum \infty e^{- 2 n}

x y^{^{'}} = y^{2} + y

What is the tangent of f(x)=4x^2-x^3,(1,3) ?
The tangent of f(x)=4x^2-x^3,(1,3) is y=5x-2