Math HL Topic: Functions, any help greatly appreciated

[Nemo]

Hi, can anyone help me with this task:

Given that the domain of the function 3x/(5+x^2) is x>=a (x is greater than or equal to a), find the least value of a such that this function has an inverse function.

The answer is supposed to be a=sqrt(5) (or 2.24), but I don't seem to understand why

Thank you very much

Edited October 6, 2016 by Nemo

[mac117]

You can use your calculator to find the turning point. An inverse exists only when each x-value corresponds to a single y-value.

When you increase x, y increases as well. After x= sqrt(5) y starts to decrease again. You can see that on the graph below.

After x=sqrt(5) the y-values start to repeat again, for different values of x, meaning that after the point where x=sqrt(5) the inverse does not exist.

You can use the horizontal line test to see when/whether the function has an inverse. If the horizontal line cuts through the function more than once, that function does not have an inverse (this works also for specific values, like in your example).

Hope that helped!

[Nemo]

@mac117 Thank you soo much!

[mac117]

7 minutes ago, Nemo said:

@mac117 Thank you soo much!

No problem  

[ultimateone]

5 hours ago, mac117 said:

You can use your calculator to find the turning point. An inverse exists only when each x-value corresponds to a single y-value.

When you increase x, y increases as well. After x= sqrt(5) y starts to decrease again. You can see that on the graph below.

After x=sqrt(5) the y-values start to repeat again, for different values of x, meaning that after the point where x=sqrt(5) the inverse does not exist.

You can use the horizontal line test to see when/whether the function has an inverse. If the horizontal line cuts through the function more than once, that function does not have an inverse (this works also for specific values, like in your example).

Hope that helped!

Is this HL only?

[kw0573]

@Nemo Because the question says sqrt (5), ie an exact value, you should learn how to do this by hand. For example, the largest point of a such that f'(a) = 0 AND (a, f(a)) is not an inflection point. 
f'(x) = (3 *(5 + x^2) - 3x(2x)) / (5 + x^2)^2 = 0

3 *(5 + x^2) - 3x(2x) = 15 + 3 x^2 - 6x^2 = 0

15 = 3 x^2

5 = x^2, the two points where f'(x) = 0 is at x = +/- sqrt(5)
Now you just have to show that x = sqrt (5) is a maximum or a minimum. If f(sqrt(5)) happened to be an inflection point, you should check each subsequent smaller roots to f'(x) = 0 until you find a max/min. You have to because y = x^3 has a point of inflection at x = 0, but the inverse function exists for all x. 

@ultiateone I can't say that it's HL only because both inverse functions and finding local maximums and minimums are in the SL syllabus. This question may appear as one of the last questions on an exam. It may be helpful to see the process of solving this question for your benefit. 

[Nemo]

@kw0573 Thank you very much!