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March 3rd 2009, 04:56 PM
Kayla_N

Related Rate

please help me...

1. Consider the function http://webwork1.math.utah.edu/webwor...28a7e6e591.png.

The absolute maximum value is: ???
and this occurs at http://webwork1.math.utah.edu/webwor...9019052b81.png???

The absolute minimum value is: ???
and this occurs at http://webwork1.math.utah.edu/webwor...9019052b81.png ??

2. Consider the function http://webwork1.math.utah.edu/webwor...cdb101f191.png.

Find the absolute minimum value of this function.
Answer: ??

find the absolute maximum value of this function.
Answer: ??

Find the absolute maximum and absolute minimum values of the function

http://webwork1.math.utah.edu/webwor...e01a88ef31.png
on the interval http://webwork1.math.utah.edu/webwor...b557e056c1.png.
Absolute maximum = ???

Absolute minimum = ????
March 3rd 2009, 05:05 PM
matheagle

In these you need to check the endpoints and where f'=0.
That's rather straight forward.
So in (1) f'=0 at x=0.
So you just need to compare f(0), f(-5) and f(2).
But clearly, since this is an 'upside down' parabola the max occurs at x=0 and that value is 4.
The min occurs at the endpoints, just comapre those two.
March 3rd 2009, 05:06 PM
mollymcf2009

Quote:

Originally Posted by Kayla_N

please help me...

1. Consider the function http://webwork1.math.utah.edu/webwor...28a7e6e591.png.

The absolute maximum value is: ???
and this occurs at http://webwork1.math.utah.edu/webwor...9019052b81.png???

The absolute minimum value is: ???
and this occurs at http://webwork1.math.utah.edu/webwor...9019052b81.png ??

2. Consider the function http://webwork1.math.utah.edu/webwor...cdb101f191.png.

Find the absolute minimum value of this function.
Answer: ??

find the absolute maximum value of this function.
Answer: ??

Find the absolute maximum and absolute minimum values of the function

http://webwork1.math.utah.edu/webwor...e01a88ef31.png
on the interval http://webwork1.math.utah.edu/webwor...b557e056c1.png.
Absolute maximum = ???

Absolute minimum = ????

To find minimum and maximum values you need to take the derivative and then use a sign chart to see where the sign of the derivative changes from - to + OR + to - on the given interval.
When the sign of f' changes, the graph of f will have a max or min.
March 3rd 2009, 05:15 PM
matheagle

In 2 you can use calculus also, but it's real easy to complete the square.
$x^4-32x^2+5=x^4-32x^2+256-256+5=(x^2-16)^2-251$ .
Hence the abs min occurs when x=-4 or 4, value is -251.
BUT you need to make sure these are x's in your region.
Well -4 isn't, so abs min occurs at x=4, now check for enpoints to get your max.
March 3rd 2009, 05:42 PM
Kayla_N

ok 3 f'(x)= (8x+4)(4)-4x(8) / (8x+4)^2

maximum= 4/9
minimum= 1/3
March 3rd 2009, 06:00 PM
matheagle

Quote:

Originally Posted by Kayla_N

ok 3 f'(x)= (8x+4)(4)-4x(8) / (8x+4)^2

maximum= 4/9
minimum= 1/3

set f'=0
and look at f at your two endpoint, ttyl