Calculus help?

Question

Find the absolute maximum and minimum values, if any, of the given function.

x
------- on [-1,3]
x^2+3

I found the derivative to be

-1
-------
x^2+3

So now I dont know how to find what numbers to test to see where it increases and decreases. My derivative could be wrong so you might check that.

I appreciate it, so thanks in advance!

Answer 1

First off, your derivative is wrong. You need to use the quotient rule. if you have (u/v), the derivative is (u'*v-u*v')/(v^2).

Second, to find critical points, set the derivative to 0 and solve for x.

Once you know the critical points, take the second derivative and evaluate it at the critical points. If the second derivative is negative, your function is "concave down" at that point and you have a maximum. If the second derivative is positive, then your function is "concave up" and you have a minimum.

Answer 2

The derivative is [(x^2 + 3) - 2x^2]/(x^2 + 3)^2 =
= (3 - x^2)/ (x^2 + 3)^2

It gets the value zero at x=sqrt(3) and -sqrt(3), but -sqrt(3) is not in [-1,3] becuase -sqrt(3)<-1

If x 3 - sqrt(3)^2 = 3-3 = 0 (ascending)

If x>sqrt(3) 3-x^2 < 3 - sqrt(3)^2 = 3-3 = 0 (descending)

Hence sqrt(3) is a local maximum points at which

sqrt(3)/(sqrt(3)^2 + 3) = sqrt(3)/6

at -1

-1/[(-1)^2 + 3] = -1/4

at 3:

3/(3^2 + 3) = 3/12 = 1/4

sqrt(3)/6 > -1/4

sqrt(3)/6 > 1.5/6 = 1/4

Hence sqrt(3) is the absolute maximum point on [-1, 3]

and -1 is the absolute minimum point.

Answer 3

The function can have its absolute maximum(minimum) at a critical point or at the extremes of the interval

y´=1/(x^2+3)^2 *(x^2+3 -2x^2)
y¨= 0 x^2=3 so x= sqrt3 (given the interval)
So we have to test
f(-1)= -1/4
f(sqrt3) = sqrt3 /6=0.2887
f(3) = 3/12 = 1/4 = 0.25
The absolute minimum is f(-1) =-1/4
the absolute maximum is f(sqrt3) =sqrt3 /6