Question

i need to practice on finding minimum/maximum can someone give me something and check my answer afterwards? it'll help a lot...

Answer 1

find the max and min of \(xe^x\) on \([-2,1]\)

Answer 2

hmm okay...so the first derivative of \(xe^x\) i know to be \[xe^x + e^x\]
if i equate to 0
\[e^x(x + 1) = 0\]
i have x = -1

what do i do with [-2. 1]?

Answer 3

you have the critical point, to find the max and min, you have to check the critical point and also the endpoint of the interval
biggest is max
smallest is min

Answer 4

try \(x^2\ln(x)\) for a global min

Answer 5

how do i "check the critical point and also the endpoint of the interval" substitute?

Answer 6

or for an easier one minimum value of \(7x+\frac{48}{x}\) on \((0,\infty)\)

Answer 7

yes substitute

Answer 8

you need to check both the critical points and also the endpoints of the interval

Answer 9

where do i substitute? in the original equation?

like in xe^x i sub in x = -1

so -1/e?

Answer 10

yes that is the minimum

Answer 11

don't forget the minimum value is the
(y\) value, not the \(x\) value
calculus find the \(x\) value to give the max and min, but min in the output not the input

Answer 12

try
\[\frac{x^2}{|x-2|}\] on \((2,5)\) for a challenge

Answer 13

so what was the maximum for that xe^x?

Answer 14

wait...just to clarify that [-2,1] was just to test if x = -1 is within the given limits right?

Answer 15

you have to check endpoints as well
at \(-2\) you get \(-2e^{-2}=-\frac{2}{e^2}\)
at \(-1\) you get \(-e^{-1}=-\frac{1}{e}\)
and at 1 you get \(e\)

Answer 16

compare the numbers you see that
\(-\frac{1}{e}\) is the smallest at \(e\) is the largest

Answer 17

if you want more, let me know

Answer 18

ohhh so e is the max and -1/e is the min?

Answer 19

yes

Answer 20

it is on the interval \([-2,1]\)

Answer 21

ahhh nice!!

Answer 22

what does global min mean?

Answer 23

minimum value over the entire domain of the function

Answer 24

ohh so no limits...i assume that's where second derivative comes in?

Answer 25

no you can use the first derivative if you like

Answer 26

or second to test whether you have a max or min

Answer 27

\[x^2 \ln x\]
\[y' = x + 2x \ln x\]
\[x(1 + 2 \ln x = 0\]
\[x = 0 \; \ln x = -\frac 12 \implies x = e^{-1/2}\]

i would say e^-1/2 is the global min yes?

Answer 28

\[x = 0 \; ; \; \ln x = -\frac 12 \implies x = e^{-1/2}\]

Answer 29

no

Answer 30

and the reason that \(e^{-\frac{1}{2}}\) is not the global minimum is that it is the \(x\) value

Answer 31

don't forget, the minimum is not the input, it is the output

Answer 32

ohh substitute...

Answer 33

right!

Answer 34

\[e^{-1/2 \times 2} \ln (e^{-1/2}) \implies e^{-1} \times -\frac 12 \implies -\frac{1}{2e}\]
right?