What is the tangent of f(x)=8x^3,\at x= 1/2 ?

The tangent of f(x)=8x^3,\at x= 1/2 is y=6x-2

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tangent of f(x)=8x^3,\at x= 1/2

tangent of

f (x) = 8 x^{3}, at x = \frac{1}{2}

y = 6 x - 2

Solution steps

Find the tangent point:

(\frac{1}{2}, 1)

Find the slope of

f (x) = 8 x^{3} : \frac{df ( x )}{d x} = 24 x^{2}

Compute the slope of

f (x) = 8 x^{3}

x = \frac{1}{2} : m = 6

Find the line with slope m=

6

and passing through

(\frac{1}{2}, 1) : y = 6 x - 2

y = 6 x - 2

arctan (cos (θ))

\frac{d y}{d x}, y = \frac{- 5 x}{5} + 3 x^{3} - \frac{4}{5 x}

y = 3 x^{2} - 5 x + 9

y^{^{''}} + 9 y = 3 sec (3 x)

5 x e^{2 x}

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