SOLUTION: I got 3 for the absolute minimum for f(x) = x^4 – 4x^3 – 5. Is that correct?
Also, if you don’t mind, could you check this one to see if I got the right answer?
The area bel
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Question 126974: I got 3 for the absolute minimum for f(x) = x^4 – 4x^3 – 5. Is that correct?
Also, if you don’t mind, could you check this one to see if I got the right answer?
The area below y = 4 – x^2 over the interval [0, 2] is 8/3. Is this correct or not?
Thank you so much!
Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website!
(3,f(3)) is at least a local minimum because the first derivitive goes to 0 at 3 and the second derivitive is positive at 3. The picture proves that the minimum is absolute rather convincingly.
2nd problem.
=
See the graph below for an illustration of the unreasonability of your answer. The secant line forms a right triangle with the axes with an area of 4. By inspection you can see that the area of this triangle is less than the desired area under the function between the given limits, yet your answer of is clearly less than 4.
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