use a graphing utility to graph ( f(x)=x^{3}-4x + 4 ) on the interval ( -2,2 ) and approximate any local…

use a graphing utility to graph ( f(x)=x^{3}-4x + 4 ) on the interval ( -2,2 ) and approximate any local maximum values and local minimum values. determine where the function is increasing and where it is decreasing. using a graphing utility, graph the function for ( -2leq xleq2 ) and ( -4leq yleq10 ). choose the correct graph. find the local maximum. select the correct choice and, if necessary, fill in the answer boxes to complete your choice. a. there are two local maxima. the leftmost maximum is and occurs at ( x = ). the rightmost maximum is and occurs at ( x = ) (round to two decimal places as needed.) b. there is one local maxima. it is and it occurs at ( x = ). (round to two decimal places as needed.) c. there is no local maximum.

use a graphing utility to graph ( f(x)=x^{3}-4x + 4 ) on the interval ( -2,2 ) and approximate any local maximum values and local minimum values. determine where the function is increasing and where it is decreasing. using a graphing utility, graph the function for ( -2leq xleq2 ) and ( -4leq yleq10 ). choose the correct graph. find the local maximum. select the correct choice and, if necessary, fill in the answer boxes to complete your choice. a. there are two local maxima. the leftmost maximum is and occurs at ( x = ). the rightmost maximum is and occurs at ( x = ) (round to two decimal places as needed.) b. there is one local maxima. it is and it occurs at ( x = ). (round to two decimal places as needed.) c. there is no local maximum.

Answer

Explanation:

Step1: Find the derivative of the function

The function is (f(x)=x^{3}-4x + 4). Using the power rule ((x^{n})^\prime=nx^{n - 1}), the derivative (f^\prime(x)=3x^{2}-4).

Step2: Find the critical points

Set (f^\prime(x)=0), so (3x^{2}-4 = 0). Then (x^{2}=\frac{4}{3}), and (x=\pm\frac{2}{\sqrt{3}}\approx\pm1.15).

Step3: Analyze the sign of the derivative

  • For (x\in[-2,-\frac{2}{\sqrt{3}})), let (x=-1.5), then (f^\prime(-1.5)=3\times(-1.5)^{2}-4=3\times2.25 - 4=6.75-4 = 2.75>0).
  • For (x\in(-\frac{2}{\sqrt{3}},\frac{2}{\sqrt{3}})), let (x = 0), then (f^\prime(0)=3\times0^{2}-4=-4<0).
  • For (x\in(\frac{2}{\sqrt{3}},2]), let (x = 1.5), then (f^\prime(1.5)=3\times(1.5)^{2}-4=3\times2.25 - 4=6.75-4 = 2.75>0).

Since the function changes from increasing to decreasing at (x =-\frac{2}{\sqrt{3}}\approx - 1.15), we find (f(-\frac{2}{\sqrt{3}})=(-\frac{2}{\sqrt{3}})^{3}-4\times(-\frac{2}{\sqrt{3}})+4=-\frac{8}{3\sqrt{3}}+\frac{8}{\sqrt{3}}+4=\frac{- 8 + 24}{3\sqrt{3}}+4=\frac{16}{3\sqrt{3}}+4\approx6.96).

Answer:

B. There is one local maxima. It is (6.96) and it occurs at (x=-1.15).