a. Locate the critical points of ƒ. b. Use the First Derivative Test to locate the local maximum and minimum values. c. Identify the absolute maximum and minimum values of the function on the given interval (when they exist). f(x)=x2x2−1 on [−4,4]

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a. Locate the critical points of ƒ.  b. Use the First Derivative Test to locate the local maximum and minimum values.  c. Identify the absolute maximum and minimum values of the function on the given interval (when they exist).  f(x)=x2x2−1 on [−4,4]

Ian Adams · Accepted Answer

Step 1Given: f(x)=x2x2−1 on the interval [−4,4]To find critical points, find x such that f′(x)=0f(x)=x2x2−1f′(x)=(x2−1)2x−x2(2x)(x2−1)2f′(x)=−2x(x2−1)2Equate f′(x) equal to 0−2x(x2−1)2=0−2x=0x=0Thus, the critical point is x=0.Step 2Now the intervals therefore will be [−4,0] and [0,4]Assign a test value for each interval to determine if Increasing or Decreasing.Intervalxf′(x)Remarks[−4,0]−249Increasing[0,]−12169Increasing[0,4]2−49DecreasingThe change from increasing to decreasing corresponds to local maximum and the change from decreasing to increasing corresponds to local minimum.Step 3Therefore, the local maximum is at x=0.(c) Solve for f(x) using the end intervals and critical pointsxf(x)Remarks−41.06AbsoluteMaximum00AbsoluteMinimum41.06AbsoluteMaximumThus, the absolute maximum is at x=−4,4 and the absolute minimum is at x=0.

Locate the critical points of ƒ. Use the First Derivative Test to locate the local maximum and minimum values. Identify the absolute maximum and minimum values of the function on the given interval. f(x)=\frac{x^{2}}{x^{2}-1} on [-4,4]

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