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

The tangent of f(x)=(x^3+2)/3 ,\at x=-3 is y=9x+56/3

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tangent of f(x)=(x^3+2)/3 ,\at x=-3

tangent of

f (x) = \frac{x ^{3} + 2}{3}, at x = - 3

y = 9 x + \frac{56}{3}

Solution steps

Find the tangent point:

(- 3, - \frac{25}{3})

Find the slope of

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

Compute the slope of

f (x) = \frac{x ^{3} + 2}{3}

x = - 3 : m = 9

Find the line with slope m=

9

and passing through

(- 3, - \frac{25}{3}) : y = 9 x + \frac{56}{3}

y = 9 x + \frac{56}{3}

\frac{d}{d x} (x^{4} + y^{4})

\frac{d}{d x} (\frac{x}{x ^{3} + 1})

f (x) = cos^{2} (2 x)

\frac{1 - 2 x}{1 + 2 x}

\frac{d}{d x} (3^{x} - 2)

What is the tangent of f(x)=(x^3+2)/3 ,\at x=-3 ?
The tangent of f(x)=(x^3+2)/3 ,\at x=-3 is y=9x+56/3