Y min and max values

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Loki123

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Sep 22, 2021
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790
The problem goes
What do you get when you multiply the mininum value of y and maximum value of y in the interval [-1,4]
I have written what y is in the pictures
My work is pretty useless but I'll post it so it's not like I didn't try. I think I have the basic idea figured, but am just missing a lot of information. What do I do? IMG_20220108_183852.jpg
 
Give your function a quick sketch. The f'(x) method only gives relative maxima and minima. It could be that the endpoints f(-1) and f(4) are higher than the relative maximum.

-Dan
 
topsquark said:
Give your function a quick sketch. The f'(x) method only gives relative maxima and minima. It could be that the endpoints f(-1) and f(4) are higher than the relative maximum.

-Dan
Click to expand...
I tried, I got 6 x 111 but that's not even in the suggested answers
 
Loki123 said:
The problem goes
What do you get when you multiply the mininum value of y and maximum value of y in the interval [-1,4]
I have written what y is in the pictures
My work is pretty useless but I'll post it so it's not like I didn't try. I think I have the basic idea figured, but am just missing a lot of information. What do I do? View attachment 30543
Click to expand...


The last line of your work has:-

25-50+15 = 40

...you might want to check this calculation :)
 
You found your local min is at [imath]x=\sqrt{5}[/imath] and your local max at [imath]x=0[/imath].
Then [imath]f(\sqrt5)=-10[/imath], [imath]f(0)=15[/imath], [imath]f(-1)=6[/imath], and [imath]f(4)=111[/imath]
What's the question asking?
 
BigBeachBananas said:
You found your local min is at [imath]x=\sqrt{5}[/imath] and your local max at [imath]x=0[/imath].
Then [imath]f(\sqrt5)=-10[/imath], [imath]f(0)=15[/imath], [imath]f(-1)=6[/imath], and [imath]f(4)=111[/imath]
What's the question asking?
Click to expand...
What do you get when you multiply min and max.
 
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