Need help finding derivatives and concavity.

In summary: Since $c$ is positive this would make $x<0$ or imaginary...is this part of the stated domain?No, this is not part of the stated domain.
  • #1
yli
10
0
Hi, I am having some trouble with this problem.
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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.
 

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  • #2
Hello, and welcome to MHB, yli! (Wave)

The quotient rule states:

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

Now, let's look at what we have:

\(\displaystyle f_1(x)=bc+ax^d\)

\(\displaystyle f_2(x)=c+x^d\)

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

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

\(\displaystyle f_2'(x)=\,?\)
 
  • #3
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?
\[g\prime(x)=\dfrac{(dx^{d-1})(bc+ax^d)-(c+x^d)(adx^{d-1})}{(c+x^d)^2}\]
 
  • #4
yli said:
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:

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

Now, factor the numerator completely...what do you get?
 
  • #5
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.
 
  • #6
yli said:
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?
 
  • #7
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?
 
  • #8
yli said:
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:

\(\displaystyle x^{d-1}=0\implies x=0\)

Is this part of the stated domain?

yli said:
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:

\(\displaystyle (c+x^d)^2=0\)

\(\displaystyle c+x^d=0\)

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

Since $c$ is positive this would make $x<0$ or imaginary...is this part of the stated domain?
 
  • #9
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?
 
  • #10
yli said:
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:

\(\displaystyle g'(x)>0\)

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

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

to:

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

Okay, earlier you stated:

yli said:
...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:

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

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

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

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

What critical values are in the stated domain?
 
  • #11
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?
 
  • #12
yli said:
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:

\(\displaystyle c(d-1)-(d+1)x^d=0\)

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

FAQ: Need help finding derivatives and concavity.

1. What is a derivative?

A derivative is a mathematical concept that describes the rate at which one variable changes in relation to another variable. It is the slope of a tangent line to a curve at a specific point.

2. How do I find the derivative of a function?

To find the derivative of a function, you can use the power rule, product rule, quotient rule, or chain rule depending on the type of function. These rules involve taking the derivative of each individual term in the function and combining them using the appropriate rule.

3. What is concavity?

Concavity is a geometric property of a curve that describes whether the curve is curving upwards or downwards at a specific point. It is determined by the second derivative of a function.

4. How do I determine the concavity of a function?

To determine the concavity of a function, you can find the second derivative of the function and then evaluate it at a specific point. If the second derivative is positive, the function is concave up at that point. If it is negative, the function is concave down. If the second derivative is zero, the function has an inflection point.

5. Why is finding derivatives and concavity important?

Finding derivatives and concavity is important in many areas of mathematics and science, including calculus, physics, and economics. It allows us to analyze the behavior of a function and make predictions about its future values. It is also essential for optimization problems, where we need to find the maximum or minimum of a function.

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