# Use first derivative test to find local minima and local maxima

Use the First Derivative Test to find the points of local maxima and minima of the function $ƒ(x)=2x^3−x^4$.

To begin we have $f'(x)=6x^2-4x^3$

Then we can set $f'(x)=6x^2-4x^3=0$ and factor it as $2x^2(3-2x)=0$

The we get $\frac{3}{2}$ as a critical point and maxima. Is $x=0$ also a critical point?

Sorry if this is an easy question I just keep having problems with this in class.

Now I have this chart but it is saying I am wrong

Function $2x^3−x^4$

------------------Intervals

Properties:

                  $(−∞,1.5)---- 1.5---(1.5,∞)  Signs of$f(x)$>>>>>>>>> + >>>>>>>>>>>>>> - Increase/decrease of$f(x)$increasing >>local max>>decreasing ## 4 Answers First Derivative Test: Find all points where$f'(x)=0$or is undefined, these are called critical points. Any of these could be an extrema. However that's not the only option. The point could also be an "inflection point". An inflection point$(a, f(a))$, is a point where the derivative$f'(a)=0$, but where$f'(a-\delta)$and$f'(a+\delta)$, where$\delta$is a very small number, have the same sign (both$f'(a-\delta)$&$f'(a+\delta)$are positive, or both are negative). An inflection point, is neither a maximum or a minimum. NOTE: This is not the technical definition of an inflection point, but it does provide a method that will work for any continuous function$f: D \subset \Bbb R \to \Bbb R$, assuming the$\delta$you choose is "small enough". Second Derivative Test: Plug in the values you found for$f'(x)=0$into the second derivative$f''(x)$. If:$\begin{cases} f''(x)\gt 0 & \text{Then f(x) is a relative minimum} \\ f''(x)\lt 0 & \text{Then f(x) is a relative maximum} \\f''(x)=0 & \text{Then the test is inconclusive} \end{cases}$So for your example:$f'(x)=6x^2-4x^3=0$when$x=0$or$x=\frac 32$. And$f'(x)$is a polynomial so it is defined over the entire domain of$f$. That means that your potential relative maxima and minima are at$(0,f(0))$and$(\frac 32, f(\frac 32))$. Let's use the second derivative test:$f''(x)=12x-12x^2=12x(1-x)=0$when$x=0$or$x=1$and is again defined everywhere. So let's figure out where it's positive and negative by choosing three points$c_1, c_2, c_3$such that$c_1 \lt 0 \lt c_2 \lt 1 \lt c_3$. How about$c_1=-1$,$c_2=\frac 12$, and$c_3=2$-- those seem like pretty easy numbers. Plugging those into$f''(x)$, we get$f''(-1)=-12-12=-24\lt 0f''(\frac 12)=6-3=3 \gt 0$and$f''(2) = 24-48=-24 \lt 0$So we've found that$f(x)$is concave up on the interval$(0,1)$and concave down on the set$(-\infty, 0)\cup(1,\infty)$. So now let's evaluate the points$x=0$and$x=\frac 32$that we found via the first derivative test. We see that$\frac 32 \in (-\infty, 0)\cup(1,\infty)$, so we know that the point$(\frac 32, f(\frac 32))$is a relative maximum. On the other hand$x=0$is inconclusive with the second derivative test. So let's check if it's an inflection point via the method I described above. Let's make our$\delta=\frac 14$and evaluate$f'(-\frac 14)$and$f'(\frac 14)$.$f'(-\frac 14)=6(\frac {1}{16})-4(\frac{-1}{64})=\frac {7}{16} \gt 0$and$f'(\frac 14)=6(\frac {1}{16})-4(\frac{1}{64})=\frac {5}{16} \gt 0$So$(0,f(0))$is an inflection point of$f(x)$and thus not an extrema. • I see so the only local extra is at$\frac{3}{2}$which is a local maxima Nov 8 '14 at 17:17 • @BS319 Yep. That's right. – user137731 Nov 8 '14 at 17:20 • so for my chart where intervals are dec and inc would look like:$(inf,1.5)$decreasing$1.5$max and$(1.5, inf)$decreasing Nov 8 '14 at 17:26 • To find out if you function$f$is increasing or decreasing you need to plug in values of$x$both above and below$1.5$into$f'(x)$. Say$x=1$and$x=2$. If you get a positive number, your graph is increasing there, if it is negative you know that it is decreasing there. Because we've found the relative extrema, inflection points, and know that this function is continuous you can then say that the sign of$f'(1)$corresponds to increasing or decreasing over the entire set$(-\infty, 0)\cup(0,1.5)$. Likewise$f'(2)$will tell you about the interval$(1.5,\infty)$. – user137731 Nov 8 '14 at 17:46 • Note that the function is not increasing or decreasing at$x=0$or$x=1.5$. As can be seen from the plot. – user137731 Nov 8 '14 at 17:47 If you insist on using the First Derivative Test, what you need to do is check whether your function is increasing or decreasing at points close to your critical points,$\frac{3}{2}$and$0$. That is, if$f'(1) > 0$and$f'(2) < 0$, you can conclude that you have a local maximum at$x = \frac{3}{2}$. You're looking for a change from increasing to decreasing or decreasing to increasing to show something is a local maximum/local minimum. You do something similar for points around$x = 0$. Also, critical points come from three places: 1. Endpoints of an interval are always critical points (If you have a given interval). 2. Places for which your derivative function is undefined (Your derivative function is defined everywhere). 3. The$x$-values where$f'(x) = 0$(Both$\frac{3}{2}$and$0$do this ). • so we also need to use intervals to show when it is increasing and decreasing. So since the domain is all real numbers I used these intervals to make a table$(-\inf, 0)$which is dec$(0,1.5)$which is inc$(1.5,\inf)$which is dec and local min at$x=0$and max at$x=1.5$but I did something wrong somewhere Nov 8 '14 at 16:08 • You should check your work again. I didn't get the same thing. Nov 8 '14 at 16:11 • Check$f'(-1)$. Nov 8 '14 at 16:15$ x_1=0 $could be a maxima, but it could also be a saddle point. To verfiy if it's a saddle point, create the second derivative - if$ f''(x_1)=0 $then it's a saddle point, if$ f''(x_1) \ne 0 $then it's a maxima. • so$f''(x) = 12x-12x^2=0$so$f''(0) = 0= 0$does that mean its a maxima? Nov 8 '14 at 15:57 • for$x1 = 0$it means it's a saddle poinnt (see my answer) for$x2 = 1.5$it means it's a maxima Nov 8 '14 at 16:13 • This is false;$f(x)=x^4$has a minimum at$0$but$f''(0)=0$. Rather, we ought to look for the first$n$such that the derivative$f^{(n)}(x)$which is not zero. If$n$is odd, it is a saddle point, if$n$is even, it is an extrema, and if no such$n$exists, such a test is inconclusive. Nov 8 '14 at 16:19 • correct, nothing to add. Nov 8 '14 at 17:14 The first derivative is positive before$x=3/2$and negative after$x=3/2$. Since it vanishes at$x=3/2$, it must be a (local) maximum point. • So now my question is about$x=0\$ does that count as a local min? Nov 8 '14 at 16:03
• No, it is an inflection point, since the function grows on both sides of that point. See the graph: wolframalpha.com/input/?i=plot+2x%5E3-x%5E4 Nov 9 '14 at 9:31