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{d}{d x} (arccos (f (x)))

\frac{d}{d x} (- 2 a sin (2 x) + b cos (2 x) \cdot 2)

x \to 0 + lim (\frac{l ( x )}{x ^{3}})

x \to 8 lim (\frac{4}{( x - 8 ) ^{2}})

\frac{d}{d x} (cos^{3} (x) + sin^{4} (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