Decreasing Intervals: x_{1} < x_{2} implies f(x_{1}) > f(x_{2})
This function is
increasing over
-oo
< x < -2 and
3 < x < oo and is decreasing over -2 < x < 3
Assume f '(x) > 0 for all x in (a,b).
I. Let f '(x) > 0 for all x in (a,b).
Let x_{1}
< x_{2}
_{
}The Mean Value Theorem says
f '(c) = f(x_{2}) - f(x_{1})
x_{2} - x_{1}
_{
}Since
f '(x) > 0 and x_{2} - x_{1}
> 0, then f(x_{2}) - f(x_{1}) > 0
Therefore
f(x_{2})
> f(x_{1}) and f(x) is increasing
II. Let f '(x) < 0 for all x in (a,b).
The proof is similar to the proof in I.
I. Assume f '(x)
changes
from negative to positive at c.
Then f '(x)
< 0 for all x in (a,c) and f
'(x) > 0 for all x in (c,b).
Thus f(x)
decreases on (a,c) and increases on (c,b).
Therefore, f(c)
is a minimum.
II. Assume f '(x) changes from
positive
to negative at c.
The proof is similar
to the proof in I.