# CALCULUS, EXTREME VALUES of function on interval and where they occur

CALCULUS, EXTREME VALUES of function on interval and where they occur!!!

find the extreme value of function on the interval and where they occur.

1. y=cos(x-(pi/4)) , at the interval of [0,7pi/4]
My attempt: y'= -sin (x-(pi/4))
y= pi/4, 5pi/4

f(0)= -1/sqrt(2) f(7pi/4)= 0 f(pi/4)= 0 f(5pi/4)=-1
Max of 1 @ x=pi/4
Min of -1 at x=5pi/4
teacher said f(0) was wrong i dont know why

2. y= 4x/(x^2+1)
My attempt: y'= -4x/(x^2+1)^2
f(0)=0
Max of 0 at x=0
dont know why i got that wrong

tnx a lot

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For f(0) it's -sin(-pi/4) = - (-sqrt2 / 2) = sqrt2 / 2 = 1/sqrt2
Could you show your work for how you got f'(x) for #2?

For f(0) it's -sin(-pi/4) = - (-sqrt2 / 2) = sqrt2 / 2 = 1/sqrt2
Could you show your work for how you got f'(x) for #2?

For f(0) the negitive of sin is later added onto -1/sqrt(2) right? making it positive? ok i see what i did.
y= 4x(x^2+1)^-1
y'=-4x(x^2+1)^-2

oh ... shoot i was supposed to use quotient rule right?

thanks alot i kind of get calculus a little bit more

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For #1, when you say f(0), you mean f(0)=cos(0-(pi/4))? Or f'(0)?

For #2, you're supposed to use the quotient rule.

yes, i mean the cos one not f'(0)
when you are looking for critical points you plug in 0 of f(0) into the original/position equation right? i think i did that right just forgot about the pos/neg stuff

HallsofIvy
Homework Helper

find the extreme value of function on the interval and where they occur.

1. y=cos(x-(pi/4)) , at the interval of [0,7pi/4]
My attempt: y'= -sin (x-(pi/4))
y= pi/4, 5pi/4
You mean x, not y.

f(0)= -1/sqrt(2) f(7pi/4)= 0 f(pi/4)= 0 f(5pi/4)=-1
Max of 1 @ x=pi/4
Min of -1 at x=5pi/4
teacher said f(0) was wrong i dont know why
f(0)= cos(-pi/4)= cos(pi/4)= 1/sqrt(2). Cosine is an even function.

2. y= 4x/(x^2+1)
My attempt: y'= -4x/(x^2+1)^2
f(0)=0
Max of 0 at x=0
dont know why i got that wrong

tnx a lot
What interval was this on? The entire real line? A continuous function does not necessarily have a max or min on such an interval. Here, it should be clear that y is positive for positive x so 0 cannot be a maximum. In any case, you have the derivative wrong. Try it again.

ok. i think i got #2 down.

2. y'= 4(x^2+1) - 4x(x)
________________
(x^2+1)^2

= 4x^2+4-8x^2
_______________
(x^2+1)^2

0 is only critical pt?
f(0) = 0

sine chart for 0
-2 = neg and 2 = neg so..... no extreme values??

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ok. i think i got #2 down.

2. y'= 4(x^2+1) - 4x(x)
________________
(x^2+1)^2

= 4x^2+4-8x^2
_______________
(x^2+1)^2

0 is only critical pt?
f(0) = 0

sine chart for 0
-2 = neg and 2 = neg so..... no extreme values??

Am i correct?? did i perform the steps right??? Thanks

SammyS
Staff Emeritus
Homework Helper
Gold Member

How is it that x = 0 is a critical point?

The derivative looks OK now.

HallsofIvy
Homework Helper

ok. i think i got #2 down.

2. y'= 4(x^2+1) - 4x(x)
________________
(x^2+1)^2

= 4x^2+4-8x^2
_______________
(x^2+1)^2

0 is only critical pt?
f(0) = 0
No, 0 is not a critical point. f'(0) exists and is not 0. The numerator of the derivative is $4- 4x^2$ which is 0 at x= 1 and x= -1. The denominator is never 0.

sine chart for 0
-2 = neg and 2 = neg so..... no extreme values??
You mean "sign", not "sine".