Critical Points and Global Extrema Question

[ardentmed]

Hey guys,

I'm doubting some of my answers and I'd appreciate some help.

I'm only asking about 2abc, ignore 1ab please:

For 1a, I simply took the derivative (as I did with all three of these questions) and calculated global and local extrema and critical points.

Ultimately, I found that f($\pi$/6) = -$\pi$/6 - √3 for the absolute minimum. Also, the absolute minimum is f(-$\pi$) = 2- $\pi$ . Is this correct? How would I go about proving these two? Should I include a first derivative test just to make sure?

Moreover, for 1b, I got a quaint answer which I'm highly doubtful of.
A got an absolute maximum of f(0)=infinity and an absolute minimum of f(1) = 0. This is because I found a critical point when assessing f ' (x) = DNE at x=0. Where did I go wrong with this question?


Finally, for the last part, 1c, I got x= +/- 1/4 and x= 1/√2 for the critical points. As for the endpoints, I computed f(-1) = -1/e and f(4) = 4/(e^16) Am I on the right track? I sense that the exponent should play a role in computing the critical points, but I don't know how exactly (perhaps it should be an f'(x)=DNE critical point). Thanks in advance.

 

[Ackbach]

For part a, the solutions to setting the derivative equal to zero all look like:
\begin{align*}
x&=\pi \left( 2n-\frac16\right), n\in\mathbb{Z} \\
x&=\pi \left( 2n+\frac76 \right), n\in\mathbb{Z}
\end{align*}
You should figure out which values of $n$ give you $x$ values in the interval $[-\pi, \pi]$.

For part b, $0$ is not in the interval of interest.

For part c, I don't think you have correctly computed the critical points. I get $x_c=\pm 1/\sqrt{2}$. Then you check those two points (since they're both in the interval) as well as the endpoints.

 

[ardentmed]

For part a, the solutions to setting the derivative equal to zero all look like:
\begin{align*}
x&=\pi \left( 2n-\frac16\right), n\in\mathbb{Z} \\
x&=\pi \left( 2n+\frac76 \right), n\in\mathbb{Z}
\end{align*}
You should figure out which values of $n$ give you $x$ values in the interval $[-\pi, \pi]$.

For part b, $0$ is not in the interval of interest.

For part c, I don't think you have correctly computed the critical points. I get $x_c=\pm 1/\sqrt{2}$. Then you check those two points (since they're both in the interval) as well as the endpoints.

Thanks for the reply.

For part c, I computed:

f(1/√2) = 0.42888 as the absolute maximum and
f(-1/√ 2) = -0.4888 as the absolute minimum.

Whereas the endpoints gave me local extrema.

Am I on the right track?

Also, could you please clarify where I did wrong in part b? Isn't x=0 a critical point?

Also, I don't quite grasp the expression you wrote for a and the purpose it is supposed to serve.

Thanks again.

 

[Ackbach]

Thanks for the reply.

For part c, I computed:

f(1/√2) = 0.42888 as the absolute maximum and
f(-1/√ 2) = -0.4888 as the absolute minimum.

Whereas the endpoints gave me local extrema.

Am I on the right track?

Note quite. Endpoints can never be local extrema! Why is that? Because a local extremum is a point that is the lowest point in some open interval containing the point. But if it's an open interval containing the point, then the interval has to have points on both sides of the local minimum. That's not possible with endpoints: for a right-hand endpoint, you can't have any values greater than the endpoint, and for a left-hand endpoint, no values lesser.

Also, could you please clarify where I did wrong in part b? Isn't x=0 a critical point?

Well, it's a point where the derivative is zero or does not exist, I grant you. However, we only need to consider such points that are in the interval of interest (or in this case, you can think of the function $f(x)=\dfrac{\ln(x)}{x^2},\; x\in [1,3]$ as a function with an artificially restricted domain.) So $x=0$ is a point where the derivative is zero or does not exist, but it's not in the interval $[1,3]$. Therefore, you can ignore it.

Also, I don't quite grasp the expression you wrote for a and the purpose it is supposed to serve.

Thanks again.

So, if you let $f(x)=x-2\cos(x),\; x\in[-\pi,\pi]$, then $f'(x)=1+2\sin(x)$ on the same interval (technically, you'd probably need to make it the open interval now $x\in(-\pi,\pi)$.) Following our usual closed-interval method, we set this derivative equal to zero, and obtain the equation
$$\sin(x)=-\frac12.$$
The expressions I wrote before are ALL solutions of this equation on the entire real line. You can verify: plug in various values of $n$ (they must be integers!), take the $\sin$ of that value, and you should get $-1/2$.

 

[ardentmed]

Alright, thanks for the insightful response.

I re-did the questions and here are my results:

For a, I computed f($\pi$/√6) to be the absolute minimum, and the absolute maximum is f($\pi$) = $\pi$ + 2

For b, I computed the absolute maximum to be f(√e) = (1/2)e^(-1)

The absolute negative value is f(1) = 0.

As for c, I computed the same values as before.

I'm confused though, why are those not viable values? Are absolute values solely the endpoints? How can that be?

Thanks again.

 

[Dethrone]

I've leave the answer to Ackbach, but it is to my understanding that an absolute extrema are the highest or lowest points of the graph, which can occur at the critical numbers, or at the endpoints. However, relative extrema do not occur at endpoints, and they also don't have to be the lowest or highest points.

 

[Ackbach]

Alright, thanks for the insightful response.

I re-did the questions and here are my results:

For a, I computed f($\pi$/√6) to be the absolute minimum,

I get that $f(-\pi/6)$ is the absolute minimum. $\pi/6$ is not a critical point. If you graph the function, you'll see what I mean. (You should always graph your function!)

and the absolute maximum is f($\pi$) = $\pi$ + 2

I agree with this.

For b, I computed the absolute maximum to be f(√e) = (1/2)e^(-1)

Correct.

The absolute negative value is f(1) = 0.

You should say, "The absolute minimum value is $f(1)=0$."

As for c, I computed the same values as before.

Yes, those are correct.

I'm confused though, why are those not viable values? Are absolute values solely the endpoints? How can that be?

See Rido12's excellent summary below.

Thanks again.

I've leave the answer to Ackbach, but it is to my understanding that an absolute extrema are the highest or lowest points of the graph, which can occur at the critical numbers, or at the endpoints. However, relative extrema do not occur at endpoints, and they also don't have to be the lowest or highest points.

 

[ardentmed]

I get that $f(-\pi/6)$ is the absolute minimum. $\pi/6$ is not a critical point. If you graph the function, you'll see what I mean. (You should always graph your function!)
I agree with this.
Correct.
You should say, "The absolute minimum value is $f(1)=0$."
Yes, those are correct.
See Rido12's excellent summary below.

But why is it negative for $f(-\pi/6)$ ? I can clearly see it in the graph, but my f'(x)=0 point came out as arcsin(1/2) which should be $f(\pi/6)$

Thanks a ton for the help.

 

[Ackbach]

But why is it negative for $f(-\pi/6)$ ? I can clearly see it in the graph, but my f'(x)=0 point came out as arcsin(1/2) which should be $f(\pi/6)$

$\begin{align*}
f(x)&=x-2\cos(x) \\
f'(x)&=1+2\sin(x) \overset{\text{set}}{=} 0 \\
0&=1+2\sin(x) \\
-1&=2\sin(x) \\
-\frac12&=\sin(x) \\
\sin^{-1}\left(-{\frac{1}{2}}\right)&=x
\end{align*}$

Thanks a ton for the help.

You're very welcome!

 

[ardentmed]

\begin{align*}
f(x)&=x-2\cos(x) \\
f'(x)&=1+2\sin(x) \overset{\text{set}}{=} 0 \\
0&=1+2\sin(x) \\
-1&=2\sin(x) \\
-\frac12&=\sin(x) \\
\sin^{-1}\left(-{\frac{1}{2}}\right)&=x
\end{align*}
You're very welcome!

Thanks for the clarification. I checked all of my answers and here are my results:

http://i.share.pho.to/d11a1ffa_o.png
http://i.share.pho.to/a4cb5658_o.png
http://i.share.pho.to/f5213c0d_o.png

Thanks in advance.