Consider the function f(x)= (3x)/((1+4x^2)^2),`-4<=x` 1) The absolute maximum value is 2) and this occurs at x equals 3) The absolute minimum value is 4) and this occurs at x equals

2 Answers | Add Yours

sciencesolve's profile pic

sciencesolve | Teacher | (Level 3) Educator Emeritus

Posted on

You should remember that you need to find the zeroes of first derivative equation to determine extreme values of the function such that:

`f'(x) = ((3x)'((1+4x^2)^2) - (3x)((1+4x^2)^2)')/((1+4x^2)^4)`

`f'(x) = (3((1+4x^2)^2) - (6x)(1+4x^2)(8x))/((1+4x^2)^4)`

You need to factor out `3(1+4x^2)`  such that:

`f'(x) = (3(1+4x^2)(1 + 4x^2 - 16x^2))/((1+4x^2)^4)`

`f'(x) = (3(1 - 12x^2))/((1+4x^2)^3)`

You need to solve f'(x) = 0, hence, since `((1+4x^2)^3)!=0` , then `1 - 12x^2 = 0`  such that:

`1 - 12x^2 = 0 =gt x^2 = 1/12 =gt x_(1,2) = +-sqrt12/12`

The function increases over interval `(-sqrt12/12 ; sqrt12/12)`  and it decreases over`(-oo,-sqrt12/12)U(sqrt12/12,oo).`

Hence, the function reaches its minimum at `x = -sqrt12/12 ` and it reaches its maximum at `x = sqrt12/12.`

lfryerda's profile pic

lfryerda | High School Teacher | (Level 2) Educator

Posted on

It looks like your bounds were cutoff in the question title, but the answer is going to be similar to what I outline below.  Since you have a closed boundary (-4 <= x <= your upper bound), you can find the absolute max and min by taking the first derivative of f(x) and setting it to zero, then solving for x.  In this case, 

f'(x) = (3(1+4x^2)^2-48x^2(1+4x^2))/(1+4x^2)^4

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

by the quotient rule.  Setting f'(x)=0, we get 1=12x^2, which means that x = root 12 or x = - root 12.  The values of x are sometimes called the critical numbers.

Now evaluate the boundaries and the critical numbers.  The largest is the absolute max and the smallest is the absolute min.

f(-4) = -12/65^2

f(-root 12) = -3 root 12 / 49^2

f( root 12) = 3 root 12 / 49^2

f( your upper bound ) = put your number in for x here.

We’ve answered 317,598 questions. We can answer yours, too.

Ask a question