Identify local extrema

Question

For any parameter a, identify local extrema of the function f(x)=ax+[(x^2)-4] ((x^2)-4 is in absolute value bars) and determine their type (max/min). State for what a does f have only one local minimum.

Answer 1

Definition extremum: A function f has a local minimum at x* if f(x)> f(x*) for some neighborhood around x*. Similarly, a function f has a local maximum at x* if f(x) < f(x*) for some neighborhood around x*.

 

We will apply the First Derivative test. This states that if f has a local extremum at x* and f is differentiable at x* then f'(x*) = 0 (the derivative of f at x* is 0). We also need to look at points where f' does not exist.

 

Preliminary 1: You will need to know how to take derivative involving absolute value. Let u be a function of x, i.e. u = u(x). If f(x) = |u| then f'(x) = u*u'/|u| where u' is derivative of u. Note you can derive this by using the fact that |u| = square root of u^(2) (take derivative and use the chain rule).

 

Using above you will find derivative of |x²-4| (let u = x²-4) is  (x²-4)*(2x)/|x²-4|

 

Preliminary 2: We will need to know where x²-4 is positive and negative. 

x²-4 is positive when x>2 or x<-2. This means |x²-4| = x²-4 on these intervals.

x²-4 is negative when -2<x<2. This means |x²-4| = -(x²-4)

 

Preliminary 3: We will need to check the points x = -2 and x=2. This is because f' does not exist at this points. Why?

Because we would get a divide by zero for the derivative of |x²-4| (see prelim 1).

 

Preliminary 4: A couple ways to determine if an extremum is a max or min. 1 way is to use the 2nd derivative test. The way I use in this problem is I use the fact that if x* is a local min then for some ε > 0,  f'(x) < 0 for all x such that x*-ε<x<x* and f'(x)>0 for all x such that x*<x<x*+ε.

Similarly if x* is a local max then f'(x) > 0 for all x such that x*-ε<x<x* and f'(x)<0 for all x such that x*<x<x*+ε.

 

Back to problem:

We have f(x) = ax + |x²-4|. Taking derivative we get

f'(x) = a + (x²-4)*(2x)/|x²-4|.

Setting f'(x) = 0 and solve for x. We do this by multiplying by |x²-4|

 

Thus we have a|x²-4|+(x²-4)*(2x) = 0.

 

Now we need to use Prelim 2.

 

Case 1: x > 2 or x < -2.

In this case |x²-4| = x²-4. Since x ≠ 2 and x ≠-2 we can divide and we are left with 

a+2x = 0. This means x = -a/2 is a critical point. Note in this case a < -4 or a > 4 by our hypothesis.

 

Case 2: -2<x<2

In this case |x²-4| = -(x²-4).  Since x ≠ 2 and x ≠-2 we can divide and we are left with

a-2x = 0. This means x = a/2 is a critical point. Note in this case -4<a<4 by our hypothesis.

 

Now lets look at different points of a.

 

1.) Assume a < -4. The critical point then x = -a/2 which means the critical point is > 2. Thus our intervals of interest are (-inf, -2), (-2, 2), (2, -a/2), (-a/2, inf)

 

a) Pick a point less than -2. This means we are in Case 1 from above. Say x = -3. We have 

f'(x) = a+2x = a-6. Since a < -4 this means f'(x) < 0.

 

b) Pick a point in between (-2, 2). This means we are in Case 2 from above. Say x = 0. We have 

f'(x) = a-2*0 = a < -4 which means f'(x) < 0.

 

a & b imply -2 is NOT a local extrema since no change of sign.

 

c) Now pick a point between (2, -a/2). This means we are in Case 1 from above. Do get a point take the average of the interval so we get a midpoint. Let x = (2+-a/2)/2 = (2-a/2)/2

 

We see 

f'(x) = a+2*{(2-a/2)/2} = a+(2-a/2) = a/2 + 2 < 0 (when doing the math). Hence f'(x) < 0.

 

b and c imply 2 is NOT a local extrema since no change of sign.

 

d) now pick a point (-a/2, inf). This means we are in Case 1 from above. Say x = -a/2+1. We see

f'(x) = a+2*(-a/2+1) = 2 > 0. Hence f'(x) > 0.

 

c) and d) imply that at x = -a/2 we have a local minimum.

 

2.) We now need to look at case -4<a<4. This is Case 2 Look at the intervals (-inf, -2), (-2, a/2), (a/2, 2), (2, inf).

Do same steps from above.

You will find x = -2 is a local minimum.

You will find x = a/2 is a local maximum.

You will find x = 2 is a local minimum.

 

3.) Now look at case a>4. This is case 1 and our intervals (-inf, -a/2), (-a/2, -2), (-2, 2), (2, inf).

Do same steps from above.

You will find x = -a/2 is a local minimum.

You will find x = -2 is NOT a local extremum.

You will find x = 2 is NOT a local extremum.

 

4.) Now you need to consider the case where a = -4 or a = 4.

In this case f'(x) never equals 0 so we need only consider intervals (-inf, -2), (-2, 2), (2, inf).

If a = -4 then x = 2 is a local minimum.

If a = 4 then x = -2 is a local minimum.

 

From #1 if a < -4 we see we have 1 local minimum. From #2, -4<a<4 we have 2 local minimums. From case 3 a>4 we have 1 local minimum. From case 4 we if a = -4 we have 1 local minimum and if a = 4 we have 1 local minimum. 

 

Thus our answer is f has one local minimum if a < -4 if a > 4, if a = -4, or if a = 4.

 

Hope this was helpful. Please message me if you have any questions or If I seemed to make a mistake anywhere.

Thanks - David

 

Answer 2

Hopefully you've already graphed this for several values of a to get an idea of how the function behaves.  Note that (x²-4) is negative for -2 < x < 2 and positive otherwise.  Therefore, we could restate f(x) as

 

ax + 4 - x²;  -2 < x < 2

ax + x² -4;   Otherwise

 

So, f'(x) = 

 

a - 2x; -2 < x < 2

a + 2x; x > 2 or x < -2

 

Also, note that f''(x) = 

 

-2;  -2 < x < 2

2;  x > 2 or x < -2

 

Setting f'(x) equal to zero, we find minimums or maximums at:

 

x = a/2;  -2 < x < 2

x = -a/2; x > 2 or x < -2

 

So, if -4 < a < 4, the first equation will give us a min or max at x = a/2.  Since f''(x) is negative at this point, x = a/2 is a local maximum.  Note that in this case the sign of f'(x) changes at x = ±2, so these are local minimum.

 

If a > 4 or a < -4, then the second equation tells us there will be a local min or max at x = -a/2.  Since the second derivative is positive in this range, this will be a local minimum.  If we consider the points x = ±2, we see that the sign of f'(x) doesn't change, so these are not max or min.

 

If a = 4, neither equation gives us a solution, but if we check the points where the derivative is undefined, then we can observe at x = -2 that the slope changes from a negative to a positive.  Therefore, x = -2 will be a minimum.  At x = 2, the slope does not change sign, so this is not a max or min.

 

Similarly, for a = -4, we will have a minimum at x = 2 only.  Note that in both these cases where a = ±4, the minimum occurs at x = -a/2.  

 

In summary, if -4 < a < 4, then there are local minimums at x = ±2 and a local maximum at x = a/s.  Otherwise, there will be  a local (and absolute) minimum at x = -a/2.

 

The range where there is a single minimum is a ≥ 4 or a ≤ -4.

 

If I lost you on any of that, please let me know.  If you haven't graphed it yet, please do so, as it will make much better sense if you graph examples of each case.