Finding Relative Extrema Mathematically: r(x) = ln(x+5)/(x2+1) over a domain of [1,4]

[Raptor]

Hello, I'm having trouble finding the relative extrema of a function mathematically for a homework question. Here it is:

r(x) = ln(x+5)/(x²+1) over a domain of [1,4].

With all the other questions I've answered, I use the following method:

1) Find the first derivative
2) Set the derivative to zero; Solve for x to find critical values
3) Plug x into original function; Solve for y to get coordinates of critical values
4) Find second derivative
5) Use second derivative test to evaluate concavity

The problems I'm facing relate to steps 2 and 4. I'm unable to solve for zero and I keep getting lost finding the second derivative.

For the first derivative, I've found the derivatives of each term and used the quotient rule and I got:


(((x²+1)/(x+5))-2xln(x+5))/(x²+1)²

I'm stuck solving for zero for the numerator: ((x²+1)/(x+5))-2xln(x+5)) =0

Plugging it into the graph, I know x should = 0.061 but I'm not sure how to get there mathematically. I know it's not necessary for the extrema because I need to know the relative min and max between x =1 and x=4 but I need to prove concavity to determine the shape of the graph.

Any help would be appreciated.

 

[Deleted member 4993]

Hello, I'm having trouble finding the relative extrema of a function mathematically for a homework question. Here it is:

r(x) = ln(x+5)/(x²+1) over a domain of [1,4].

With all the other questions I've answered, I use the following method:

1) Find the first derivative
2) Set the derivative to zero; Solve for x to find critical values
3) Plug x into original function; Solve for y to get coordinates of critical values
4) Find second derivative
5) Use second derivative test to evaluate concavity

The problems I'm facing relate to steps 2 and 4. I'm unable to solve for zero and I keep getting lost finding the second derivative.

For the first derivative, I've found the derivatives of each term and used the quotient rule and I got:


(((x²+1)/(x+5))-2xln(x+5))/(x²+1)²

I'm stuck solving for zero for the numerator: ((x²+1)/(x+5))-2xln(x+5)) =0

Plugging it into the graph, I know x should = 0.061 but I'm not sure how to get there mathematically. I know it's not necessary for the extrema because I need to know the relative min and max between x =1 and x=4 but I need to prove concavity to determine the shape of the graph.

Any help would be appreciated.

Is it:

\(\displaystyle \displaystyle{r(x) = \frac{ln(x+5)}{x^2+1}}\)

or

\(\displaystyle \displaystyle{r(x) = ln \left[\frac{x+5}{x^2+1}\right ]}\)

 

[Raptor]

Hello, it is the first one Subhotosh, thank you.

 

[JeffM]

Hello, I'm having trouble finding the relative extrema of a function mathematically for a homework question. Here it is:

r(x) = ln(x+5)/(x²+1) over a domain of [1,4].

With all the other questions I've answered, I use the following method:

1) Find the first derivative
2) Set the derivative to zero; Solve for x to find critical values
3) Plug x into original function; Solve for y to get coordinates of critical values
4) Find second derivative
5) Use second derivative test to evaluate concavity

The problems I'm facing relate to steps 2 and 4. I'm unable to solve for zero and I keep getting lost finding the second derivative.

For the first derivative, I've found the derivatives of each term and used the quotient rule and I got:


(((x²+1)/(x+5))-2xln(x+5))/(x²+1)²

I'm stuck solving for zero for the numerator: ((x²+1)/(x+5))-2xln(x+5)) =0

Plugging it into the graph, I know x should = 0.061 but I'm not sure how to get there mathematically. I know it's not necessary for the extrema because I need to know the relative min and max between x =1 and x=4 but I need to prove concavity to determine the shape of the graph.

Any help would be appreciated.

First off, you do not need the second derivative. You can simply calculate r(x) at each point where r'(x) = 0. However you need to remember that, when finding extrema in an interval, you must test the endpoints as well. So you also need to compute r(1) and r(4).

Your first derivative looks ok to me. Clearing fractions and moving to exponentials does not give a nice clean equation to work with.

\(\displaystyle \dfrac{x^2 + 1}{x + 5} - 2xln(x + 5) = 0 \implies x^2 + 1 - (2x^2 + 10x)ln(x + 5) = 0 \implies\)

\(\displaystyle x^2 + 1 = (2x^2 + 10x)ln(x + 5) = \ln \left ( (x + 5)^{(2x^2 + 10x)} \right ) \implies ln \left ( e^{(x^2+1)} \right ) = \ln \left ( (x + 5)^{(2x^2 + 10x)} \right ) \implies\)

\(\displaystyle e^{(x^2+1)} = (x + 5)^{(2x^2 + 10x)}.\)

There may be a clever way to solve that beast, but I do not immediately see it. There may not be a closed form solution. So I might try the Newton-Raphson method.