What is the tangent of f(x)=8x(x+2),\at x=3 ?

The tangent of f(x)=8x(x+2),\at x=3 is y=64x-72

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tangent of f(x)=8x(x+2),\at x=3

tangent of

f (x) = 8 x (x + 2), at x = 3

y = 64 x - 72

Solution steps

Find the tangent point:

(3, 120)

Find the slope of

f (x) = 8 x (x + 2) : \frac{df ( x )}{d x} = 8 (2 x + 2)

Compute the slope of

f (x) = 8 x (x + 2)

x = 3 : m = 64

Find the line with slope m=

64

and passing through

(3, 120) : y = 64 x - 72

y = 64 x - 72

\int cot^{3} (x) csc^{5} (x) d x

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

\frac{\partial}{\partial x} (x^{4} sin (y^{9} z^{9}))

What is the tangent of f(x)=8x(x+2),\at x=3 ?
The tangent of f(x)=8x(x+2),\at x=3 is y=64x-72