1. The problem statement, all variables and given/known data
a)f(x) = sqrt(3 + x^2) - 3 x, text( ) text( ) [4, 6] [Find the max. and min. values]
b)y = (3 - x)/(x^2 + 9 x), [2, 7]
3. The attempt at a solution
a)I know how to get the max. and min. value, but before knowing them I need to know how to get the critical points from f(x). The derivative of f(x) is 2x/(sqrt(2+x^2) - 3. I don't know how to get critical points when dealing with fractions.b) I know the derivative of y, then again I don't know how to get critical points when dealing with fractions.
EDIT: In y = x^2 #8211; 6x #8211; 4, [-4, 0], why is the minimum -4 and not -13?
-13 is from the critical point 3, -4 is from the interval 0.
Originally Posted by asz304 1. The problem statement, all variables and given/known data
a)f(x) = sqrt(3 + x^2) - 3 x, text( ) text( ) [4, 6] [Find the max. and min. values]
b)y = (3 - x)/(x^2 + 9 x), [2, 7]
3. The attempt at a solution
a)I know how to get the max. and min. value, but before knowing them I need to know how to get the critical points from f(x). The derivative of f(x) is 2x/(sqrt(2+x^2) - 3.
Check your math. You have two errors in your derivative.
Why do you have text() text() in your problem statement?
For critical points, set f'(x) = 0, and solve that equation.
Since you have a restricted domain, be sure to check both endpoints, because it's possible that a max or min value occurs at either endpoint. Originally Posted by asz304 I don't know how to get critical points when dealing with fractions.b) I know the derivative of y, then again I don't know how to get critical points when dealing with fractions.
EDIT: In y = x^2 – 6x – 4, [-4, 0], why is the minimum -4 and not -13?
-13 is from the critical point 3, -4 is from the interval 0.
So in a) it's x/(sqrt(3+x^2)), the critical point is 0?
b)From y = (3 - x)/(x^2 + 9 x) on [ 3 ; 10 ]the derivative is (x^2 -6x -27)/ (x^2+9x)^2, so it's critical points are -9,3, and 0? So, f(-9) = infinity, f(3) = 0, f(0) = infinity. Then with f(10) = -7 / 190. The computer ( webassign.net ) told me that minimum is -1/27 and not -7/190.....but I got 0 as the maximum which is right.From my edit part in my previous post : In y = x^2 -6x - 5 [-1,0] , the derivative has a critical point of 3 and f(3) = -13 , f(-1) = 2, f(0) = -5, when I had my answers plugged in on webassign.net I had the wrong answers if I said -13 is the minimum and 2 is the maximum then I checked the right answer it gave me -5 as the minimum and 2 as the maximum..... why is -5 the minimum?Question: If I have an open interval or an interval that isn't restricted, should I never check if they have maximum or minimum values? Thanks
Originally Posted by asz304 So in a) it's x/(sqrt(3+x^2)), the critical point is 0?
No and no.
f'(x) = x/sqrt(3 + x2) - 3
0 is not a critical point. Originally Posted by asz304 b)From y = (3 - x)/(x^2 + 9 x) on [ 3 ; 10 ]the derivative is (x^2 -6x -27)/ (x^2+9x)^2, so it's critical points are -9,3, and 0?
Your derivative is correct on this one. The function is not defined at x = 0 and x = -9, so those aren't critical points. Factor the numerator to find the two critical points. The max and min will occur either at the two critical points or at the endpoints of the interval. Originally Posted by asz304 So, f(-9) = infinity, f(3) = 0, f(0) = infinity. Then with f(10) = -7 / 190. The computer ( webassign.net ) told me that minimum is -1/27 and not -7/190.....but I got 0 as the maximum which is right.From my edit part in my previous post : In y = x^2 -6x - 5 [-1,0] , the derivative has a critical point of 3 and f(3) = -13 , f(-1) = 2, f(0) = -5, when I had my answers plugged in on webassign.net I had the wrong answers if I said -13 is the minimum and 2 is the maximum then I checked the right answer it gave me -5 as the minimum and 2 as the maximum..... why is -5 the minimum?
f'(x) = 0 when x = 3, but 3 is not in the interval [-1, 0]. BTW, f(3) = 32 - 6(3) - 5 = 9 - 23 = -14, not -13 as you have.
Since there is no critical number in the given interval, any maximum or minimum must occur at an endpoint of the domain.
f(-1) = (-1)2 - 6(-1) - 5 = 2
f(0) = 02 - 6(0) - 5 = -5
Originally Posted by asz304
Question: If I have an open interval or an interval that isn't restricted, should I never check if they have maximum or minimum values? Thanks
If I understand what you're asking, if the function is defined on an open interval or the entire number line, then there aren't any endpoints to the interval, so obviously you don't need to check endpoints (there aren't any).
The derivative of a) sqrt(3 + x^2) - 3 x is x/sqrt(3 + x2) - 3, I used chain rule and 2 is cancelled on both denominator and numerator. So when setting f'(x) = 0, I end up having 3 sqrt( x^2 + 3) = x, how should I continue?
b) A similar and simpler example to b) is f(x) = (1 - x) / ( x^2 + 3x), it's derivative has a numerator of( x^2 - 2x -3), the factor is x = -3 and x = 1 and the closed interval is [1,6] since x = -3 is not on the closed interval the only critical point on the closed interval is 1 ( Thanks to your reply I now understand the difference of close and open intervals ). So when I was getting the maximum and minimum values, I used f(1) = 0, f(6) = -5/54, but for some reason when I checked the right answer in the computer for the minimum value it gave -1/9 and said my input quot; - 5 / 54 quot; is wrong.....Did I do something wrong again?
Originally Posted by asz304 The derivative of a) sqrt(3 + x^2) - 3 x is x/sqrt(3 + x2) - 3, I used chain rule and 2 is cancelled on both denominator and numerator. So when setting f'(x) = 0, I end up having 3 sqrt( x^2 + 3) = x, how should I continue?
Square both sides. Haven't you ever solved equations with radicals in them before? Originally Posted by asz304
b) A similar and simpler example to b) is f(x) = (1 - x) / ( x^2 + 3x), it's derivative has a numerator of( x^2 - 2x -3), the factor is x = -3 and x = 1
The numerator is x2 -2x - 3, as you said, but this factors into (x - 3)(x + 1). What values of x make the numerator zero? They are NOT x = -3 and x = 1. Originally Posted by asz304 and the closed interval is [1,6] since x = -3 is not on the closed interval the only critical point on the closed interval is 1 ( Thanks to your reply I now understand the difference of close and open intervals ). So when I was getting the maximum and minimum values, I used f(1) = 0, f(6) = -5/54, but for some reason when I checked the right answer in the computer for the minimum value it gave -1/9 and said my input quot; - 5 / 54 quot; is wrong.....Did I do something wrong again?
For this function, the minimum value occurs at a critical point, not an endpoint.
Originally Posted by Mark44 . Haven't you ever solved equations with radicals in them before?
Thanks, I did do radicals, but I'm having a rough day...cramming for my midterm in Chemistry.
Anyway, thanks again. |