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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
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

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myininaya
 one year ago
Best ResponseYou've already chosen the best response.2Well we know we need to find f'(t) is f(t)=2/t+10 or f(t)=2/(t+10)?

myininaya
 one year ago
Best ResponseYou've already chosen the best response.2just so i can know to check your derivative

myininaya
 one year ago
Best ResponseYou've already chosen the best response.2\[f(t)=\frac{2}{t}+10 \\ f(t)=2t^{1}+10 \\\] do you know how to find f'?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yes! It's \[f'(t)= \frac{ 2 }{ t^2 }\] am I right?

myininaya
 one year ago
Best ResponseYou've already chosen the best response.2yes 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)

myininaya
 one year ago
Best ResponseYou've already chosen the best response.2so you just need to consider the endpoints of your interval given

myininaya
 one year ago
Best ResponseYou've already chosen the best response.2tell me if you can find the absolute min?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Ohhh I understand. How can I find the absolute max. and min. ?

myininaya
 one year ago
Best ResponseYou've already chosen the best response.2there 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

myininaya
 one year ago
Best ResponseYou've already chosen the best response.2dw:1434748634077:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Oh yeah! So I plug in 4 into the original function and that will be my abs. min. ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So that means it is 10.5 ?

myininaya
 one year ago
Best ResponseYou've already chosen the best response.2no calculus was actually needed on this problem

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0If I did not known what the graph looked like would I still be able to find it?

myininaya
 one year ago
Best ResponseYou've already chosen the best response.2I 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

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@myininaya thank you so much!!!

myininaya
 one year ago
Best ResponseYou've already chosen the best response.2Means there is no abs max at all.* Since there were no critical numbers then there are no relative max/min
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