Let f(x) = 8(x - 6)^{\frac{2}{3}} + 8. A. Find all critical values. B. Find the intervals where...

Question:

Let {eq}f(x) = 8(x - 6)^{\frac{2}{3}} + 8 {/eq}.

A. Find all critical values.

B. Find the intervals where {eq}f(x) {/eq} is decreasing and increasing.

C. Find the values of x for all local maxima and minima.

Application of Differentiation(Concavity)

{eq}\text{If the function f (x) has a critical point then y =} f '(x) = 0.\\ \text{The function f (x) is increasing on those intervals where y =} f '(x) > 0.\\ \text{The function f (x) is decreasing on those intervals where y =} f '(x) < 0.\\ {/eq}

Answer and Explanation:

{eq}\text{For finding critical points of f, we have to first find out the first derivative of the function and equate it to 0.}\\ \text{For finding points of inflection of f, we have to find out the second derivative of the function and equate it to 0.}\\ f(x) = 8(x - 6)^{\frac{2}{3}} + 8\\ \text{A) for finding the critical points}\\ {/eq}

$$\text{First derivative of f i.e.}\\ f'=\frac{16}{3}(x - 6)^{\frac{-1}{3}}\\ f'=\frac{16}{3(x - 6)^{\frac{1}{3}}}\\ \text{Equate it to 0 to find all the critical points}\\ \frac{16}{3(x - 6)^{\frac{1}{3}}}=0\\ \text{Equate it to 0 to find all the critical points}\\ \text{The function is not defined on x=6}\\ \text{So, the function f have one critical point i.e. }x=6\\ $$

{eq}\text{B) for finding the intervals where f is increasing/ decreasing}\\ {/eq}

$$\text{For interval }(6,\infty), f' \text{ is positive. That means in this interval the function is increasing}\\ \text{For interval }(-\infty,6), f' \text{ is negative. That means in this interval the function is decreasing}\\ $$

{eq}\text{C) for local maxima and minima find out the second derivative of the function f}\\ {/eq}

$$\text{Second derivative of f i.e.}\\ f''=\frac{-16}{9}(x - 6)^{\frac{-4}{3}}\\ \text{As }f''\text{ is not defined on x=6. So there will be no extremas}\\ $$


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