2 questions. Function g(x) and a set of values of x issue.

[SolCon]

2 questions. Function g(x) and a "set of values of x" issue.

Hi to all.

I've got 2 questions which are small but I just need to get somethings cleared in them.
Alright.

Q.1) The function g is defined by g:x ---> x-3(sqrt.x) for x>=0.
Solve the equation g(x)=10.

For this, I tried 2 things. First, I tried to square both sides in at attempt to make it a quadratic equation and factorize it. I got this:

10 = x-3(x^0.5)
(10)^2 = [x-3(x^0.5)]^2 (squaring both sides)
100=x^2 -3x -3x + 9x^0.25
0=x^2 -6x + 9x^0.25 -100

I'm stuck here. Don't know how to proceed. :(

Another thing I did was to multiply both sides by x:

10x = x^2-3(x^2)^1/2
10x = x^2 - 3x
x^2 = 10x+3x
x = 13x/x
x= 13

This, however, is not the correct answer.

So what do I have to do here?

Q.2) State the set of values of x for which x^3-3x^2-9x+27 is a decreasing function of x.
Given values of x are x=-1 and x=3.

I used (-inf,-1) and (3,+inf) to try get the values (inf=infinity).
I used -2 for -inf and 4 for +inf in the above equations but got positive values as answers, whereas the question says that it is a decreasing function of x.

How do I get the set of values of x?

Thanks for any help with these 2 questions. :)

 

[Mentallic]

I can already see a few errors in your algebraic manipulations.

$$\left(x^{1/2}\right)^2=x\neq x^{1/4}$$

$$x\left(x^{1/2}\right)=x^{1+1/2}=x^{3/2}\neq \left(x^2\right)^{1/2}=x$$

 

[SolCon]

Thanks for pointing out those mistakes. :)

So we'll get: 100 = x^2 - 6x^3/2 + 9x
Is that right?

If so, then how do we proceed? Also, any thoughts about the second question?

 

[HallsofIvy]

Hi to all.

I've got 2 questions which are small but I just need to get somethings cleared in them.
Alright.

Q.1) The function g is defined by g:x ---> x-3(sqrt.x) for x>=0.
Solve the equation g(x)=10.

For this, I tried 2 things. First, I tried to square both sides in at attempt to make it a quadratic equation and factorize it. I got this:

10 = x-3(x^0.5)
(10)^2 = [x-3(x^0.5)]^2 (squaring both sides)
100=x^2 -3x -3x + 9x^0.25
0=x^2 -6x + 9x^0.25 -100

First, isolate the square root: $10- x= -3x^{0.5}$, then square both sides
$100- 20x+ x^2= 9x$ so $x^2- 29x+100=0$. Solve that quadratic equation.

I'm stuck here. Don't know how to proceed. :(

Another thing I did was to multiply both sides by x:

10x = x^2-3(x^2)^1/2
10x = x^2 - 3x

$x(x^{1/2})= x^{2/2}x^{1/2}= x^{3/2}$, not x.

x^2 = 10x+3x
x = 13x/x
x= 13


This, however, is not the correct answer.

So what do I have to do here?

As I said above, isolating the square root then squaring results in the quadratic equation $x^2- 29x+ 100= 0$ which, surprisingly, is easy to factor.

Q.2) State the set of values of x for which x^3-3x^2-9x+27 is a decreasing function of x.
Given values of x are x=-1 and x=3.

3x^2- 6x- 9= 3(x^2- 2x- 3)= 3(x- 3)(x+ 1)

I used (-inf,-1) and (3,+inf) to try get the values (inf=infinity).
I used -2 for -inf and 4 for +inf in the above equations but got positive values as answers,[/quote]
I have no idea what you mean by "used -2 for -inf and 4 for +inf". 2 and 4 are NOT particularly large! However, it is true that x has "turning points" at x= -1 and x= 3 so those -2 and 4 are beyond all turning points. The given function has value $(-2)^3- 3(-2)^2- 9(-2)+ 27= -8+ 6+ 18+ 27= 43$ and $(4)^3- 3(4^2)- 9(4)+ 27= 64- 48- 3+ 27= 40$.
I'm not sure how you would find the turning points without using Calculus. The derivative is $3x^2- 6x- 9= 3(x- 3)(x+ 1)$ which is negative for x between -1 and 3 and so the function is decreasing between -1 and 3.

whereas the question says that it is a decreasing function of x.

You may have misread this. The problem was to find the interval on which the function is a decreasing function.

How do I get the set of values of x?

Thanks for any help with these 2 questions. :)

 

[SolCon]

Thank you for that nice explanation for both questions. :)

About Q.1, I've got the x-values now which are x=4 and x=25. Just one thing though. The answer says that these are actually 5 or -2 which are both the square root of the x-values. Why is this? Is it because of the fact there is 3-(x)^0.5 in the question?

About Q.2, well the only problem is in the way the answer is supposed to be written and which value has to be greater than or less than x. I once stated this question in another forum and was told that we were supposed to use the (-inf,-x) (x,+inf) method. We could take the next value coming after each of the x-values. In this case, the next value after -1 is -2 which means that x would be less than -1 (x<-1), and the next value after 3 is 4 which means that x would be greater than 3 (x>3). I would then have to apply both these "other" x-values in the equation and check to see if the answer would come in negative. If it did, and the question was asking "for which so and so is a decreasing function of x", then that would mean that the "other" x-values were correct. Hence, I'd take the set of values of x as : 3<x<-1.

Whether this method is right or not, so far it has worked until I stumbled upon this particular question. It asked for the decreasing function but the "other" x-values (-2 and 4 here) gave a positive solution, not satisfying the question's demand for a "decreasing function". This made me assume that if whenever the question asked for a decreasing function and the "other" x-values gave positive results, I'd just switch the < > signs for the set of values of x.

I'm pretty sure that whatever I've said is mostly wrong and can be said to be purely relying on guess work.

If you think my assumptions are correct, well and good. But if you think they are wrong, then please don't hesitate to tell me that there is a better (and correct ) way of doing it. Basically I just need to get the set of values of x in the correct order of < and >.

 

[eumyang]

Thank you for that nice explanation for both questions. :)

About Q.1, I've got the x-values now which are x=4 and x=25. Just one thing though. The answer says that these are actually 5 or -2 which are both the square root of the x-values. Why is this? Is it because of the fact there is 3-(x)^0.5 in the question?

5 or -2 are not correct. Sounds like typos. Also, are you sure that x = 4 is a solution? Have you heard of "extraneous solutions?"

 

[SolCon]

eumyang, I have actually not heard of this term. Thanks for informing me of this.

A quick google search shows that these solutions are those that do not satisfy the original equation.

Here, like you have said, x=25 is not the extraneous solution but x=4 is.
This is what I've done:

>With x=25)

> 10 = x-3(x^0.5)
> 10 +3(x^0.5) = x
> 10 +3(25^0.5) = 25
> 10 +3(5) = 25
> 10 + 15 = 25
> 25 = 25

>But with x=4)

> 10 = x-3(x^0.5)
> 10 +3(x^0.5) = x
> 10 +3(4^0.5) = 4
> 10 +3(2) = 4
> 10 + 6 = 4
> 16 = 4

Hence, it is an extraneous solution.

However, in all the questions I've attempted thus far, there has been no mention of extraneous solutions in the answer given at the back, no mention of it any question I've done and no difference in rewarding marks for a question which may have had one. This is probably one reason why I've never heard of this term. Regardless, I thank you for this bit of information and hope that you can enlighten me on such things whenever possible.

I am also relieved to hear that the answer given in the book could be a typo (very likely as few others are also) since this makes it a little less difficult to keep track of values when handling quadratics and functions.

I'm certain the first question is now fully solved. I just need a little bit help with the second one.