SOLUTION: At what value(s) of x does f(x)= x^4-8x^2 have a relative minimum? (A) 0 and -2 only (B) 0 and 2 only (C) 0 only (D) -2 and 2 only (E) -2,0, and 2

I assume you are taking calculus:

If a function is continuous its relative maximums or minimums they
always occur either at points where the derivative is 0 or undefined.

(Note: But just because the derivative is 0 or undefined for a certain value of x, 
that does not always mean that there is a relative maximum or minimum there.)

We find the derivative of f(x)




Set 



Factor out 



Factor the expression in parentheses:



Set each factor = 0

 gives 

 gives 

 gives 

So we have 3 critical values.  Now we can find out

whether each of these is (1) a relative maximum, (2) a relative minimum,
or (3) a horizontal inflection points.

by either of two methods, the first derivative test, or
the second derivative test.  The second derivative test
is the easier, but it has the drawback that it sometimes
fails.  The first derivative test is harder, but it never
fails.

First derivative test.  The intervals to use are the open
intervals bounded by the critical values -2, 0 and 2:


Interval    |  (-oo,-2)  |  (-2,0)  |   (0,2)  | (2,oo)
Test value  |     -3     |    -1    |     1    |   3
Sign of f'  |      -     |     +    |     -    |   +
Conclusion  | decreasing |increasing|decreasing|increasing

At x = -2 the function changes from decreasing to increasing;
therefore there is a relative minimum at the value x=-2.

At x = 0 the function changes from increasing to decreasing;
therefore there is a relative maximum at the value x=0.

At x = 2 the function changes from decreasing to increasing;
therefore there is a relative minimum at the value x=2.

So the answer is "f(x) has relative minimums at x = -2 or +2
only, choice (D)
------------------------------------------------------------

The eaier way is to the second derivative test. We find the
second derivative:





Substitute the critical values in f''(x)

  It's positive, so there is a relative minimum there. 
  It's negative, so there is a relative maximum there.
  It's positive, so there is a relative minimum there.

Notice this was an easier method, and gives the same answer.

When the second derivative is positive the curve is concave upward, and
thus we have a minimum.

When the second derivative is negative the curve is concave downward, and
thus we have a maximum.

But when the second derivative turns out to be 0, the test fails, 
and we must go to the harder first derivative test.

Edwin