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

• golf747

#### golf747

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 don't 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?

murmillo said:
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 a lot 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

golf747 said:
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 don't 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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golf747 said:
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

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

The derivative looks OK now.

golf747 said:
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".