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

1. Nov 10, 2011

### 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 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

Last edited: Nov 10, 2011
2. Nov 10, 2011

### murmillo

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

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?

3. Nov 11, 2011

### golf747

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

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

Last edited: Nov 11, 2011
4. Nov 11, 2011

### SMA_01

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

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.

5. Nov 11, 2011

### golf747

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

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

6. Nov 12, 2011

### HallsofIvy

Staff Emeritus
Re: CALCULUS, EXTREME VALUES of function on interval and where they occur!!!

You mean x, not y.

f(0)= cos(-pi/4)= cos(pi/4)= 1/sqrt(2). Cosine is an even function.

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.

7. Nov 14, 2011

### golf747

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

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??

Last edited: Nov 14, 2011
8. Nov 16, 2011

### golf747

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

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

9. Nov 16, 2011

### SammyS

Staff Emeritus
Re: CALCULUS, EXTREME VALUES of function on interval and where they occur!!!

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

The derivative looks OK now.

10. Nov 17, 2011

### HallsofIvy

Staff Emeritus
Re: CALCULUS, EXTREME VALUES of function on interval and where they occur!!!

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.

You mean "sign", not "sine".