MHB Need help finding derivatives and concavity.

[yli]

Hi, I am having some trouble with this problem.
View attachment 7483
I have completed part a but I am stuck on part b and c.
I used the quotient rule to try and find the first derivative, but I am unsure if I have done so correctly. This is my work for part b so far.
\[g\prime(x)=\dfrac{(bc+ax^d)(c+dx^{d-1})-(c+x^d)(bc+adx^{d-1})}{(c+x^d)^2}
\]
If I am doing this properly, I am a bit unsure of how I should simplify this derivative. Thanks, for any help.

 

[MarkFL]

Hello, and welcome to MHB, yli! (Wave)

The quotient rule states:

$$\frac{d}{dx}\left(\frac{f_1(x)}{f_2(x)}\right)=\frac{f_1'(x)f_2(x)-f_1(x)f_2'(x)}{f_2^2(x)}$$

Now, let's look at what we have:

$$f_1(x)=bc+ax^d$$

$$f_2(x)=c+x^d$$

Bearing in mind that $a,\,b,\,c,\,d$ are all constants, can you find:

$$f_1'(x)=\,?$$

$$f_2'(x)=\,?$$

 

[yli]

Since bc is a constant, I am assuming that it will become 0, and so the derivative of the first part should be \[(adx^{d-1})\] Then the derivative of the second part I am assuming is \[(x^{d-1})\] Sorry I am really bad at this and therefore still somewhat confused. Would this be correct?
\[g\prime(x)=\dfrac{(dx^{d-1})(bc+ax^d)-(c+x^d)(adx^{d-1})}{(c+x^d)^2}\]

 

[MarkFL]

Since bc is a constant, I am assuming that it will become 0, and so the derivative of the first part should be \[(adx^{d-1})\] Then the derivative of the second part I am assuming is \[(x^{d-1})\] Sorry I am really bad at this and therefore still somewhat confused. Would this be correct?
\[g\prime(x)=\dfrac{(dx^{d-1})(bc+ax^d)-(c+x^d)(adx^{d-1})}{(c+x^d)^2}\]

You have the negative of the derivative...you should have:

$$g'(x)=\frac{\left(adx^{d-1}\right)\left(c+x^d\right)-\left(bc+ax^d\right)\left(dx^{d-1}\right)}{\left(c+x^d\right)^2}$$

Now, factor the numerator completely...what do you get?

 

[yli]

Oh, that was a silly mistake. Once I factored I got
\[g\prime(x)=\dfrac{(a-b)cdx^{d-1}}{(c+x^d)^2}\]

I also tried to find the second derivative from this as well, would it be

\[g\prime\prime(x)=\dfrac{(a-b)cd(c+x^d)(x^{d-2}((d-1))(c+x^d)-2dx^d))}{(c+x^d)^4}
\]
I am a bit confused about how I can find the critical values for part b, and the inflection points for part c.

 

[MarkFL]

Oh, that was a silly mistake. Once I factored I got
\[g\prime(x)=\dfrac{(a-b)cdx^{d-1}}{(c+x^d)^2}\]

Yes, that's what I got as well. (Yes)

Before we move on to concavity, let's identify our critical values...that is, those values of $x$ in the given domain where either the numerator is zero, or the denominator is zero. What do you find?

 

[yli]

Hmmm, this is the part I was confused about because, critical points are where the function is equal to 0. When I set the function to 0, I am lost about what I am supposed to do next,

\[\dfrac{(a-b)cdx^{d-1}}{(c+x^d)^2}=0
\]
\[(a-b)cdx^{d-1}=0
\]
I thought about dividing (a-b) from both sides, but if I do that, I am unsure where to go from there. In this case would the critical value be at x=0? For getting the second critical value where the denominator is equal to zero, when I finish, I think I get \[−^d\sqrt{c}
\]
Does that make sense?

 

[MarkFL]

Hmmm, this is the part I was confused about because, critical points are where the function is equal to 0. When I set the function to 0, I am lost about what I am supposed to do next,

\[\dfrac{(a-b)cdx^{d-1}}{(c+x^d)^2}=0
\]
\[(a-b)cdx^{d-1}=0
\]
I thought about dividing (a-b) from both sides, but if I do that, I am unsure where to go from there. In this case would the critical value be at x=0?

Yes, since we are given $b<a$, then we know $0<a-b$, and so we are not potentially dividing by zero. $c$ and $d$ are post positive, and so they cannot be zero, so we may divide them out as well, and we are left with:

$$x^{d-1}=0\implies x=0$$

Is this part of the stated domain?

For getting the second critical value where the denominator is equal to zero, when I finish, I think I get \[−^d\sqrt{c}
\]
Does that make sense?

Yes:

$$(c+x^d)^2=0$$

$$c+x^d=0$$

$$x=(-c)^{\frac{1}{d}}$$

Since $c$ is positive this would make $x<0$ or imaginary...is this part of the stated domain?

 

[yli]

I believe that these two critical points do not exist on the graph for this function since the domain is restricted to values of x greater than 0. Also because the values of x have to be greater than 0, I am assuming that this function is always increasing. If this is correct, how would I go about finding concavity?

 

[MarkFL]

I believe that these two critical points do not exist on the graph for this function since the domain is restricted to values of x greater than 0. Also because the values of x have to be greater than 0, I am assuming that this function is always increasing. If this is correct, how would I go about finding concavity?

Yes, neither critical value is in the stated domain, and you are also correct in that for all $x$ within the stated domain, we have:

$$g'(x)>0$$

The function $g(x)$ increases monotonically from:

$$\lim_{x\to0^{+}}g(x)=b$$

to:

$$\lim_{x\to\infty}g(x)=a$$

Okay, earlier you stated:

...I also tried to find the second derivative from this as well, would it be

\[g\prime\prime(x)=\dfrac{(a-b)cd(c+x^d)(x^{d-2}((d-1))(c+x^d)-2dx^d))}{(c+x^d)^4}
\]

Let me check your result:

$$g''(x)=\frac{(a-b)cd(d-1)x^{d-2}\left(c+x^d\right)^2-(a-b)cdx^{d-1}\left(2\left(c+x^d\right)dx^{d-1}\right)}{\left(\left(c+x^d\right)^2\right)^2}$$

$$g''(x)=\frac{(a-b)cdx^{d-2}\left(c+x^d\right)\left((d-1)\left(c+x^d\right)-2dx^d\right)}{\left(c+x^d\right)^4}$$

Yes, we seem to agree here. Let's further simplify:

$$g''(x)=\frac{(a-b)cdx^{d-2}\left(c+x^d\right)\left(c(d-1)-(d+1)x^d\right)}{\left(c+x^d\right)^4}$$

What critical values are in the stated domain?

 

[yli]

Sorry, this is where I am stuck. The algebra is starting to confuse me a bit more so than usual. Could I get a hint of how I am supposed to find the critical values?

 

[MarkFL]

Sorry, this is where I am stuck. The algebra is starting to confuse me a bit more so than usual. Could I get a hint of how I am supposed to find the critical values?

From our earlier analysis of the first derivative, we know the only factor in either the numerator or denominator we need to consider is:

$$c(d-1)-(d+1)x^d=0$$

What do you get when solving for $x$? Are there any restrictions on any of the parameters we need to apply?