# Math Help - [SOLVED] Family of functions

1. ## [SOLVED] Family of functions

Given the family of functions h(x)=x+cos(ax), where a is a positive constant such that 0<a<4.

For what values of a wiill h(x) have a relative maximum at x=1?

I found the h'(x) first. Then I equated that to zero: h'(x)=1-asin(ax).
Then I'm not sure what to do next, since there are two variables, do I just plug x=1 into the equation then solve for a?

2. Originally Posted by chukie
Given the family of functions h(x)=x+cos(ax), where a is a positive constant such that 0<a<4.

For what values of a wiill h(x) have a relative maximum at x=1?

I found the h'(x) first. Then I equated that to zero: h'(x)=1-asin(ax).
Then I'm not sure what to do next, since there are two variables, do I just plug x=1 into the equation then solve for a?
Yes.

But remember that the solutions you get for a force stationary points at x = 1 and so might include values that give relative minimum and stationary points of inflexion at x = 1 .... You need to test the nature of the solutions .....

Also, you won't be able to solve $\sin a = \frac{1}{a}$ for exact values of a so you'll have to content yourself with approximate solutions ......

3. Originally Posted by mr fantastic
Yes.

But remember that the solutions you get for a force stationary points at x = 1 and so might include values that give relative minimum and stationary points of inflexion at x = 1 .... You need to test the nature of the solutions .....

Also, you won't be able to solve $\sin a = \frac{1}{a}$ for exact values of a so you'll have to content yourself with approximate solutions ......
Thanks! I've managed to find a by graphing it on my calculator. Now the next question asks me what values of a will h(x) have an inflection pt at x=1. I have found the second derivative for h(x), equated that to zero, then plug 1 in again to solve for a. I graphed it again to find the values for a, except there are no zeroes when I graphed it. Am I doing something wrong?

4. Originally Posted by chukie
Thanks! I've managed to find a by graphing it on my calculator. Now the next question asks me what values of a will h(x) have an inflection pt at x=1. I have found the second derivative for h(x), equated that to zero, then plug 1 in again to solve for a. I graphed it again to find the values for a, except there are no zeroes when I graphed it. Am I doing something wrong?
$h''(x) = -a^2 \cos (ax)$. The equation $0 = -a^2 \cos a$ has plenty of solutions.

You should be able to get exact solutions for a without using technology.

Note: $h''(1) = 0$ is a necessary but not sufficient condition for an inflexion point at x = 1. So you'll need to test all solutions to this equation.

5. Originally Posted by mr fantastic
$h''(x) = -a^2 \cos (ax)$. The equation $0 = -a^2 \cos a$ has plenty of solutions.

You should be able to get exact solutions for a without using technology.

Note: $h''(1) = 0$ is a necessary but not sufficient condition for an inflexion point at x = 1. So you'll need to test all solutions to this equation.
Ah! I see, I did the second derivative wrong. One last question, what values of a make h(x) strictly decreasing.

I was thinking that if it is strictly decreasing, the function would be monotonic, so h'(x)<0. Thus I wrote 1-asin(ax)<0, but now what value of x should I plug in? Is it still 1?

6. Originally Posted by chukie
Ah! I see, I did the second derivative wrong. One last question, what values of a make h(x) strictly decreasing.

I was thinking that if it is strictly decreasing, the function would be monotonic, so h'(x)<0. Thus I wrote 1-asin(ax)<0, but now what value of x should I plug in? Is it still 1?
If the question is interpretted as h(x) is decreasing at x = 1, then you require $h'(1) < 0 \Rightarrow 1 - a \sin a < 0$.

7. Originally Posted by mr fantastic
If the question is interpretted as h(x) is decreasing at x = 1, then you require $h'(1) < 0 \Rightarrow 1 - a \sin a < 0$.
The question doesn't say that though; it just says which values of a makes h(x) strictly decreasing. If I don't interpret it as h(x) is decreasing at x = 1, can the question be done?

I was trying out different graphs on my calculator with a. I have noticed that when 0<a<1 the graph seems to be monotonic, but increasing. I don't think there is a value for a that makes h(x) decreasing. But this is just a guess.

8. Originally Posted by chukie
The question doesn't say that though; it just says which values of a makes h(x) strictly decreasing. If I don't interpret it as h(x) is decreasing at x = 1, can the question be done?

I was trying out different graphs on my calculator with a. I have noticed that when 0<a<1 the graph seems to be monotonic, but increasing. I don't think there is a value for a that makes h(x) decreasing. But this is just a guess.
A necessary condition is that y = h(x) does not have any stationary points => values of a such that $h'(x) \neq 0 \Rightarrow 1 - a \sin (ax) \neq 0 \Rightarrow a \sin (ax) \neq 1$ for all x. So a necessary condition is that $0 < a < 1$.

However, this could mean that y = h(x) is decreasing OR increasing ......

y = h(x) is decreasing => values of a such that $h'(x) < 0 \Rightarrow 1 - a \sin (ax) < 0 \Rightarrow a \sin (ax) > 1$ for all x.

But it's impossible to satisfy this condition AND $a \sin (ax) \neq 1$.

So the answer is that there's no values of a for which h(x) is decreasing for all values of x. Does the question give an interval over which h(x) is to be strictly decreasing?

9. Originally Posted by mr fantastic
Does the question give an interval over which h(x) is to be strictly decreasing?
Nope.

10. Originally Posted by chukie
Nope.
Then there's no value of a.

11. Awesome! Thanks so much for your help!