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

The tangent of f(x)=7x^2-x^3,(1,6) is y=11x-5

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tangent of f(x)=7x^2-x^3,(1,6)

tangent of

f (x) = 7 x^{2} - x^{3}, (1, 6)

y = 11 x - 5

Solution steps

Find the slope of

f (x) = 7 x^{2} - x^{3} : \frac{df ( x )}{d x} = 14 x - 3 x^{2}

Compute the slope of

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

(1, 6) : m = 11

Find the line with slope m=

11

and passing through

(1, 6) : y = 11 x - 5

y = 11 x - 5

\int 18 x^{2} - 14 x + 9 d x

x + \frac{2}{x}

t^{2} y^{^{''}} - 2 t y^{^{'}} + 2 y = 3 t^{2}

\int_{1}^{3} arctan (\frac{1}{x}) d x

What is the tangent of f(x)=7x^2-x^3,(1,6) ?
The tangent of f(x)=7x^2-x^3,(1,6) is y=11x-5