closed interval method: problem 2\n(1 point)\nconsider the function ( f(x)=4 - 7x^{2}, - 3leq xleq1 ).\nthe…
closed interval method: problem 2\n(1 point)\nconsider the function ( f(x)=4 - 7x^{2}, - 3leq xleq1 ).\nthe absolute maximum value is\nand this occurs at ( x ) equals\nthe absolute minimum value is\nand this occurs at ( x ) equals\nnote: you can earn partial credit on this problem.\npreview my answers submit answers\nyou have attempted this problem 0 times.\nyou have unlimited attempts remaining.\nemail instructor
Answer
Explanation:
Step1: Find the derivative of the function
The function is (f(x)=4 - 7x^{2}). Using the power rule ((x^{n})^\prime=nx^{n - 1}), the derivative (f^\prime(x)=-14x).
Step2: Find the critical points
Set (f^\prime(x) = 0), so (-14x=0), which gives (x = 0). The critical point (x = 0) lies within the interval ([-3,1]).
Step3: Evaluate the function at the critical point and endpoints
- For (x=-3): (f(-3)=4-7\times(-3)^{2}=4 - 63=-59).
- For (x = 0): (f(0)=4-7\times0^{2}=4).
- For (x = 1): (f(1)=4-7\times1^{2}=4 - 7=-3).
Answer:
The absolute maximum value is (4) and this occurs at (x) equals (0). The absolute minimum value is (-59) and this occurs at (x) equals (-3).