What is the tangent of f(x)=9+5x^2-2x^3,(1,12) ?

The tangent of f(x)=9+5x^2-2x^3,(1,12) is y=4x+8

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tangent of f(x)=9+5x^2-2x^3,(1,12)

tangent of

f (x) = 9 + 5 x^{2} - 2 x^{3}, (1, 12)

y = 4 x + 8

Solution steps

Find the slope of

f (x) = 9 + 5 x^{2} - 2 x^{3} : \frac{df ( x )}{d x} = - 6 x^{2} + 10 x

Compute the slope of

f (x) = 9 + 5 x^{2} - 2 x^{3}

(1, 12) : m = 4

Find the line with slope m=

4

and passing through

(1, 12) : y = 4 x + 8

y = 4 x + 8

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

\int_{2}^{9} \frac{4}{8 x - 9} d x

\int_{1}^{4} (2 x - 2 x) d x

f (x) = 7 + 4 x^{2} - 2 x^{3}, (1, 9)

\int \frac{2 x}{- 5 - 6 x ^{2} - x ^{4}} d x

What is the tangent of f(x)=9+5x^2-2x^3,(1,12) ?
The tangent of f(x)=9+5x^2-2x^3,(1,12) is y=4x+8