## anonymous one year ago Help would be much appreciated ^.^ -Find the absolute and local maximum and minimum values of f(t) = 2/t + 10 , 0 < t ≤ 4

1. myininaya

Well we know we need to find f'(t) is f(t)=2/t+10 or f(t)=2/(t+10)?

2. anonymous

the 1st one :)

3. myininaya

just so i can know to check your derivative

4. myininaya

ok cool

5. myininaya

$f(t)=\frac{2}{t}+10 \\ f(t)=2t^{-1}+10 \\$ do you know how to find f'?

6. anonymous

Yes! It's $f'(t)= \frac{ -2 }{ t^2 }$ am I right?

7. myininaya

yes and f' is not zero ever but f' dne at t=0 but we f(0) also doesn't exist at t=0 we actually have a vertical asymptote |dw:1434748364786:dw| so we don't have to worry about critical numbers but this just shows we also don't have any max of any type but w do have a min and it is an absolute min (you know since local min can't occur at endpoints)

8. myininaya

so you just need to consider the endpoints of your interval given

9. myininaya

tell me if you can find the absolute min?

10. anonymous

Ohhh I understand. How can I find the absolute max. and min. ?

11. myininaya

there is a vertical aysmptote in the pic and as x approaches it from the right the y values keep getting bigger and bigger there is no max you only need to find the absolute min and I gave you a big hint for that it occurs at the endpoint for this question

12. myininaya

|dw:1434748634077:dw|

13. anonymous

Oh yeah! So I plug in 4 into the original function and that will be my abs. min. ?

14. myininaya

yes

15. anonymous

So that means it is 10.5 ?

16. myininaya

yes

17. myininaya

no calculus was actually needed on this problem

18. anonymous

If I did not known what the graph looked like would I still be able to find it?

19. myininaya

Yes.

20. myininaya

I said no calculus was needed because g(x)=1/x is a pretty basic graph learned in algebra. f(x)=2/x+10 is just a translation of that graph. Anyways f(x)=2/x+10 , 0<x<=4 We already know this as a vertical asymptote at x=0 And we know in the interval (0,4] the y values are approaching positive infinity since as the x's move closer to 0 since $\lim_{x \rightarrow 0^+}(\frac{2}{x}+10)=\infty$ Means there is no max at all. You could find f' to see if there are any critical numbers on (0,4) but we did that and there are none. So f will have no local min. But you can also say since $f'(t)=\frac{-2}{t^2}<0 \text{ means } f \text{ is decreasing }$ You have definitely ensure f(4) is a absolute min

21. anonymous

@myininaya thank you so much!!!

22. myininaya

Means there is no abs max at all.* Since there were no critical numbers then there are no relative max/min

Find more explanations on OpenStudy