Example 0.1. Let f (x) be the function
f (x) = x2 − 9x + 18 .
Find the critical points of f (x) and identify the regions on which f (x) is increasing or decreasing.
We have a few things to worry about this time. Notice that f (x) may not
be differentiable at the places where x2 − 9x + 18 = 0. Everywhere else, f (x) is
always either equal to x2 − 9x + 18 or equal to −x2 + 9x − 18. This means that
f ′ (x) is always either equal to 2x − 9 or −2x + 9; in either case, the derivative
is zero at x = 29 .
As stated above, there may be differentiability issues at places where x2 −
9x + 18 = 0; this happens when x = 3 or when x = 6 by factorization. It can
be easily checked that f (x) is not differentiable at either of these points. To see
this, notice that the derivatives 2x − 9 and −2x + 9 do not agree at x = 3 or at
x = 6. Therefore, 3 and 6 are critical numbers for f (x).
So f (x) has three critical points: x = 3, x = 9/2, and x = 6. Let’s plug
in some sample values between each of these points (for example, you can use
x = 0, x = 4, x = 5, and x = 7). Let’s make our table.
Region
x<3
3 < x < 9/2 9/2 < x < 6
x>6
f ′ (x)
<0
>0
<0
>0
f (x)
decreasing
increasing
decreasing
increasing
This shows that f (x) has local minima at x = 3 and x = 6, and a local
maximum at x = 9/2. The local minima of f (x) are f (3) = 0 and f (6) = 0.
The local maximum of f (x) is f (9/2) = 9/4.
The graph of this function can be seen in the geogebra file. Note the corners
at (3, 0) and at (6, 0).
Even without the picture, it should be clear that f (x) is nonnegative- it’s
an absolute value.
Let’s do another example problem for which the function fails to be differentiable. This time, let’s use that neat little trick from Wednesday.
If you find the method used in this problem confusing, don’t worry about it.
You won’t be tested on it. I just think it’s neat.
Example 0.2. Find and classify the critical points of
f (x) = ln(1 + |x − 3|).
This function is defined everywhere because 1 + |x − 3| is a strictly positive
function. As hinted above and on Wednesday, the chain rule tells us where to
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look for critical points- we have a critical point for any x for which 1+|x−3|
=0
(this never happens) or for any x at which 1+|x−3| has a critical point- and the
only such place is at x = 3, where the function is not differentiable. Therefore,
f (x) has only one critical point, at x = 3.
Note that we managed to find the critical points without ever actually computing the derivative. We don’t have immediate access to f ′ , but we only need
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to know if it’s positive or negative. The 1+|x−3|
factor is always positive (this
follows from the fact that ln(x) is increasing) and the derivative of 1 + |x − 3| is
clearly negative for x < 3 and positive for x > 3, so the critical point at x = 3
is a local minimum. Furthermore, the value of f at this point is f (3) = 0.
Note that you can use this technique to quickly find the critical points of really nasty functions. Again, If you find this technique confusing, don’t overthink
it. Just stick to the methods that you know.
p
Example 0.3. Find the critical points of f (x) = 4 ln(x2 + 5).
There’s a domain consideration here- we need to know that x2 + 5 isp
always
at least 1 so that ln(x2 + 5) is always nonnegative (it is). Furthermore, 4 ln(x)
has nonzero derivative and is always increasing, so the only place there could
possibly be a critical point is when the derivative of x2 + 5 is zero, which only
happens at x = 0. So we can quickly identify the critical points of f (x) without
taking the
√ extremely annoying derivative. Note that this strategy only worked
because 4 ln(x) is an increasing function of x.
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