What is the tangent of f(x)=x^3,\at x=4 ?

The tangent of f(x)=x^3,\at x=4 is y=48x-128

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tangent of f(x)=x^3,\at x=4

tangent of

f (x) = x^{3}, at x = 4

y = 48 x - 128

Solution steps

Find the tangent point:

(4, 64)

Find the slope of

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

Compute the slope of

f (x) = x^{3}

x = 4 : m = 48

Find the line with slope m=

48

and passing through

(4, 64) : y = 48 x - 128

y = 48 x - 128

\frac{\partial}{\partial x} (\frac{y x}{1 + y x})

\int \frac{1}{( x ^{2} + 36 ) ^{2}} d x

4 t^{3} - 24 t + 24 sin (t)

x \to 1 lim (- f (x))

\int sec (x) tan (x) + 4 x d x

What is the tangent of f(x)=x^3,\at x=4 ?
The tangent of f(x)=x^3,\at x=4 is y=48x-128