mathcalculus 2 years ago H-E-L-P so frustrating!! Determine the extrema of below on the given interval f(x)=5x^3-61x^2+16x+3 (a) on [0,4] The minimum is ?? and the maximum is ?? (b) on [-9,9] The minimum is ?? and the maximum is ?? please solve so i can know which answer i got wrong.

1. mathcalculus

can someone please tell me the correct answer to this. it seems like one of my answers is incorrect.

2. satellite73

check $$f(0), f(4)$$ and also find the critical point and find $$f$$ of that number

3. mathcalculus

a) on [0,4] The minimum is -589 and the maximum is 4.06074 (b) on [-9,9] The minimum is -8727 and the maximum is -1149

4. satellite73

the derivative is $$15 x^2-122 x+16$$

5. satellite73

which, by some miracle factors as $$(x-8) (15 x-2)$$

6. satellite73

zeros of the derivative are $$\frac{2}{15}$$ and $$8$$

7. satellite73

so $$\frac{2}{15}$$ is the $$x$$ coordinate of the local max and $$8$$ is the $$x$$ coordinate of the local min

8. mathcalculus

so for the intervals [0,4] max is: 2/15 and 8 is the min? what about [-9,9]

9. mathcalculus

i just want to make sure because it submitted the answer and it keeps saying its wrong.

10. mathcalculus

@satellite73

11. mathcalculus

any one! this is important!

12. mathcalculus

never mind! i found it. thanks to those who actually helped.

13. Peter14

f'(x)=10x^2 - 122x +16 0=10x^2 - 122x + 16 (just working here because I don't have paper handy)

14. satellite73

careful here the max is the $$y$$ value not the $$x$$ value

15. Peter14

just finding critical points

16. Peter14

oh, sorry you already found them

17. satellite73

8 is not in the interval $$[0,4]$$ for that one, you need to check $$f(0), f(4), f(\frac{2}{15})$$ the largest is the max and the smallest is the min

18. satellite73

for $$[-9,9]$$ you need to check $f(-9),f(\frac{2}{15}), f(8), f(9)$

19. mathcalculus

got it. thank you!

20. satellite73

yw