What are the minimum and maximum values?

Finding absolute maximum and minimum values?

• What is the absolute maximum and minimum values of f(x)= x^4-98x^2+4 in the closed interval [-6,15]? My answer for the maximum is f(15)=28579 which is right but my answer for the minimum, being f(-6)= -2228, is wrong. The only critical number I could find was 49^(1/3), but it can't be the absolute minimum since f(49^(1/3))= -1129.0 (approx.). The derivative is f '(x)=4(x^3-49). I think my problem might be that I have not found all the real roots of f '(x). Help!

• Answer:

  Relative or local extremes occur where f’(x) = 0 f'(x) = 4x³ - 196x = 0 so 4x(x² - 49) = 0 That is x = 0 or x = 7 or x = -7 are possible extremes (i.e., min or max) f(0) = 4, f(7) = f(-7) = -2397 since the function is an even function Since x⁴will dominate f(x), the function’s value will be very positive for very negative and also very positive x’s. So, only global minimums can occur over all. You were correct in identifying the value of the function at the end points of the domain. f(-6) = -2228, f(15) = 28,579 From the data identified, the minimum on the interval [-6, 15] occurs when x = 7 and the maximum occurs when x = 15. There is also a local maximum at x = 0 as identified using the 2nd derivative test (not shown).