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

Algebra ->  Rational-functions -> 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      Log On


   

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) About Me  (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.

graph%28600%2C600%2C-5%2C5%2C-40%2C40%2Cx%5E4-4x%5E3-5%29

2nd problem.

int%284-x%5E2%2Cdx%2C0%2C2%29=%284%282%29-%282%5E3%2F3%29%29-%284%280%29-%280%5E3%2F3%29%29=8-8%2F3=16%2F3

See the graph below for an illustration of the unreasonability of your answer. The secant line y=-2x%2B4 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 8%2F3 is clearly less than 4.