What is the tangent of f(x)=8x^2-x^3,(1,7) ?

The tangent of f(x)=8x^2-x^3,(1,7) is y=13x-6

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tangent of f(x)=8x^2-x^3,(1,7)

tangent of

f (x) = 8 x^{2} - x^{3}, (1, 7)

y = 13 x - 6

Solution steps

Find the slope of

f (x) = 8 x^{2} - x^{3} : \frac{df ( x )}{d x} = 16 x - 3 x^{2}

Compute the slope of

f (x) = 8 x^{2} - x^{3}

(1, 7) : m = 13

Find the line with slope m=

13

and passing through

(1, 7) : y = 13 x - 6

y = 13 x - 6

(x - 5)^{\frac{7}{3}}

x \to 3 lim (\frac{2}{( x - 3 )})

x^{2} + 4 x - 1

\frac{\partial}{\partial x} (2 x y + 5 y^{2} - 3 x y^{3})

x \to 1 lim (e^{5 x^{4} - 5 x})

What is the tangent of f(x)=8x^2-x^3,(1,7) ?
The tangent of f(x)=8x^2-x^3,(1,7) is y=13x-6