Today, we had a set of problems as usual, and a quiz! (And I didn't tell you about the quiz, even though others did, so I'm going to pretend that it was a pop quiz)!. Below, you'll find the three problems, their solutions, and a worked-out quiz.

We had three questions in recitation.

- A function $f(x)$ is continuous on $[1,5]$ and differentiable on $(1,5)$. We happen to know that $f'(x) > 2$.
- What can you conclude about the function from the Mean Value Theorem?
- Use the Mean Value Theorem to show that $f(5) > f(1) + 8$.

- Consider the following function $f(x) = \begin{cases} x^2 + 1 &
\text{if } x < -2 \ x^3 + 13 & \text{if } -2 \leq x < 0 \ x^4 + 13 & \text{if } 0 < x < 1 \ -14x + 28 & \text{else } \end{cases}$.
- Without using derivatives, determine where $f(x)$ is increasing or decreasing.
- Check by taking derivatives.
- Identify the local maxes and mins of the function $f(x)$.

- Consider the function $g(x) = 3x^5 - 25x^3 + 60x + 2$.
- Show that $-2, -1, 1, 2$ are the critical points of $g(x)$.
- What are the local minima and maxima of $g(x)$?
- Compute $g''(x)$ for each of the critical points of $g(x)$.
- Do you notice a pattern between the second derivatives at the minima and maxima?

I'll consider the quiz after these problems.

#### Question 1

This question is designed around the mean-value theorem. The mean-value theorem states that if you have a function $f(x)$ that is continuous on an interval $[a,b]$ and differentiable on $(a,b)$. Then there is a $c$ in the interval $(a,b)$ such that $f'(c) = \dfrac{f(b) - f(a)}{b-a}$. In other words, the "average slope" gets hit by the derivative.

So in this problem, from the mean value theorem, we know that $\dfrac{f(5) - f(1)}{5-1} = f'(c)$ for some $c$ in $(1,5)$. We know in addition that $f'(x) > 2$ always. So in fact, the mean value theorem tells us that $\dfrac{f(5) - f(1)}{5-1} = f'(c) > 2$, or that $f(5) - f(1) > 2\cdot (4)$. Rewriting this, we get that $f(5) > 8 + f(1)$, which is exactly what we were trying to show.

#### Question 2

Without using derivatives!

We'll consider each function-piece in turn. First, we have $x^2 + 1$ on
$(-\infty, -2)$. The thing that changes with $x$ here is $x^2$, and the size of $x^2$ depends only on the magnitude of $x$:
bigger input magnitude yields bigger output magnitude. So as $x$ is
increasing from $-\infty$ to $-2$, it is *decreasing* in magnitude,
and thus $x^2$ is decreasing. In a different way, we know that $x^2$ is a parabola with vertex above $x = 0$. It's decreasing for all
negative numbers and increasing on all positive numbers. So we have one part
down.

Next, we have $x^3 + 13$ on $[-2, 0)$. Our previous thought process
doesn't quite work now. $x^3$ will take negative numbers to more negative
numbers, and the larger the input (in magnitude), the larger the output. So as
$x$ is increasing from $-2$ to $0$, it is decreasing in
magnitude. This means that $x^3$ will be *increasing*, as the negative
numbers coming out will also be decreasing in magnitude. Thus $x^3$ is
increasing here. In a different way, we know the graph of $x^3$, and it's
always increasing.

$x^4 + 13$ behaves a lot like a parabola. Now we're in positive numbers,
which match our intuition a lot better than negative numbers. As the magnitude
of $x$ increases, so does the magnitude of $x^4$. So this function is
also *increasing.*

Finally, we have a line. Thank goodness, a line! This line has negative slope, so it's decreasing always.

Let's check using derivatives. The derivative of $x^2 + 1$ is $2x$, and we're only looking at negative $x$. Thus $2x$ is always negative, and so the original function is decreasing. The derivative of $x^3 + 13$ is $3x^2$. This is always nonnegative, so the original function is increasing. The derivative of $x^4 + 13$ is $4x^3$. We're now on positive $x$, so $4x^3$ is always positive, and so the original function is increasing. Finally, the derivative of our line $-14x + 28$ is $-14$, which is negative, and so the original function is decreasing. So we were right above - good.

When we are trying to identify the local maxima and minima, it's tempting to just try to use (what you're about to learn:) the first derivative test: finding the derivatives and setting equal to zero. But that won't work here. Notice that all the derivative we found are zero only at $0$. But our function isn't differentiable everywhere. Instead, we should use the fact that we now know when the function is increasing and decreasing. A local maximum will occur when the function increases, and then decreases. A local minimum will occur when the function decreases, then increases.We know that the function is decreasing until $x = -1$, after which the function is increasing. So there is a local min at $x = -2$. The function goes from increasing to decreasing at $x = 1$ too. So we know that $x = 1$ will be a local maximum. We have found all the times when the function is changing from increasing to decreasing, or from decreasing to increasing, so we're done.

As an aside:This is really the intuition behind the first derivative test, too. The idea of the first derivative test is that maxes and mins will occur when the derivative is zero or on the boundary, so long as our function is differentiable. Why is this the case? If our function is differentiable, then the slope of the function changes from positive to negative (i.e. the function changes from increasing to decreasing) when the derivative is zero. But here, our function isn't differentiable everywhere, so we need to be a bit wittier.

#### Question 3

So we are now looking at $g(x) = 3x^5 - 25x^3 + 60x + 2$. Let's differentiate: we find that $g'(x) = 15x^4 - 75x^2 + 60$. If we plug in $x = -2, -1, 1, 2$, we get zero. Yay!. In fact, we have that $g'(x) = 15(x+2)(x-2)(x+1)(x-2)$. These are the places where the derivative is zero, so we expect them to be our max and min candidates. By either plugging in numbers or looking at when the function is increasing/decreasing, or by realizing that this is a positive quintic with 4 extrema (if this makes sense - awesome; if not - don't worry), we have maxes at $x = -2, 1$ and mins at $x = -1, 2$.

The second derivative is $g''(x) = 60x^3 - 150x$. The idea behind the
second half of this problem is to motivate the second derivative test (which
you'll learn shortly). It just so happens that when you compute the second
derivative at these four points, it's negative at the two maxima and positive
at the two minima. This might lead you to make the conjecture that at local
maxima, the second derivative is always negative; and at local minima, the
second derivative is always positive. But *this would be wrong*.

The converse, however, is true. If the second derivative is negative at a critical point, then that point is a local maximum. If it's positive at a critical point, then that point is a local minimum.

Why is this true? The second derivative $g''$ is the

first derivativeof the first derivative. This means that when the second derivative is positive, the first derivative is increasing. And when it's negative, the first derivative is decreasing. So at a critical point, i.e. when the first derivative is zero, if the second derivative is positive, this means that the first derivative is increasing. Since the first derivative is zero and increasing, this means that it was just negative and is about to be positive. Butthismeans that the original function was decreasing and then increasing, i.e. that there is a local min. So when the second derivative is positive at a critical point, we get a local min. This type of reasoning works for when the second derivative is negative at a critical point too. I encourage you to try it!

I'll post up the quiz and its solutions shortly, but in a separate post. This one is very long as is, I think. I'd like to finish by reminding you all that I will not be in MRC next week. In addition, something was left in the classroom. If you're missing something and you think you left it, I am holding on to it - so let me know. (I'll give it to some lost and found office if I don't hear anything).

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Comments (1)2012-11-01 confusedHey David,

I have a question about #23. It says to graph $y=x+sin x$ for $[0 \leq x \leq 2 \pi ]$

The answer in the back of the book shows an absolute max at $(2 \pi , 2 \pi )$.

I am confused because I found $y'=1+cos x$

Therefore, to find the $0$, I did $cos x=-1$. And $cos=-1$ at $\pi$.

So...why is the absolute max at $2\pi$?

Comment 2: On 2012-11-02 davidlowryduda

This is a very good question. You are right that $y' = 1 + cos x$, so from the first derivative test we know that there is a critical point at $\pi$.

But the key difference is that we are on a particular interval: $0 \leq x \leq 2 \pi$, The first derivative test finds extrema on the inside of the interval, but the boundary must be checked separately.

So we check $x = 0$ and $x = 2\pi$ too. This means we have three candidates for global extrema: $x = 0, \pi, 2\pi$. At $0$, the function takes the value $0$. At $\pi$, the function takes the value $\pi - 0 = \pi$. At $2\pi$, the function takes the value $2 \pi$. From this, we find that the absolute max is at $x = 2 \pi$ and the absolute minimum is at $x = 0$, for $x$ in the interval $[0, 2\pi ]$.

This is a good thing to point out: In general, to maximize a function on a domain, we try the first derivative test, and check against the boundaries of the domain.

And before you doubt the utility of the first derivative test, let's see what happens if we expand the interval to, say, $[0, 5 \pi]$. From $y' = 1 + cos x$, we know that the 'suspected' critical points are at $\pi, 3\pi, 5\pi$, that is when $cos x = -1$. Now $x = 2 \pi$ is not a candidate, and it's not even a local min or max! It's only because it was at the boundary that it had the chance to be a min or max.