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

The tangent of f(x)=5+4x^2-2x^3,(2,5) is y=-8x+21

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tangent of f(x)=5+4x^2-2x^3,(2,5)

tangent of

f (x) = 5 + 4 x^{2} - 2 x^{3}, (2, 5)

y = - 8 x + 21

Solution steps

Find the slope of

f (x) = 5 + 4 x^{2} - 2 x^{3} : \frac{df ( x )}{d x} = - 6 x^{2} + 8 x

Compute the slope of

f (x) = 5 + 4 x^{2} - 2 x^{3}

(2, 5) : m = - 8

Find the line with slope m=

- 8

and passing through

(2, 5) : y = - 8 x + 21

y = - 8 x + 21

\int \frac{1}{30 - x ^{2}} d x

\int cos^{5} (\frac{3 x}{2}) sin^{7} (\frac{3 x}{2}) d x

\int - 6 \cdot sin (2 x) + 3 \cdot cos (3 x) d x

\frac{d}{d x} (sin^{9} (cos (8 x)))

What is the tangent of f(x)=5+4x^2-2x^3,(2,5) ?
The tangent of f(x)=5+4x^2-2x^3,(2,5) is y=-8x+21