What is the tangent of f(x)=(x^2-8)^5,\at x=3 ?

The tangent of f(x)=(x^2-8)^5,\at x=3 is y=30x-89

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tangent of f(x)=(x^2-8)^5,\at x=3

tangent of

f (x) = (x^{2} - 8)^{5}, at x = 3

y = 30 x - 89

Solution steps

Find the tangent point:

(3, 1)

Find the slope of

f (x) = (x^{2} - 8)^{5} : \frac{df ( x )}{d x} = 10 x (x^{2} - 8)^{4}

EN : T i tl e G e n er a l Eq u a t i o n Sl o p e A t P o in t 2 Eq : m = 30

Find the line with slope m=

30

and passing through

(3, 1) : y = 30 x - 89

y = 30 x - 89

x \to 5.6 lim (7.3 x + 8.4)

\int_{0}^{\frac{1}{2}} arccos (x) d x

(x, y) \to (0, 0) lim (\frac{y}{x})

f (x) = (x^{2} - 48)^{4}

\frac{d}{d x} (6 x^{\frac{1}{3}})

What is the tangent of f(x)=(x^2-8)^5,\at x=3 ?
The tangent of f(x)=(x^2-8)^5,\at x=3 is y=30x-89