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$f(x)$ is a real valued function defined as

$$f(x) = 1 + x + (14x-3-8x^2)^{0.5}$$

$$g(x) = \big(f(x)\big)^{0.5}$$

What is the domain of $g(1 + x)$?

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Can you show us what you have tried? – Henry Feb 23 '12 at 8:44
i created f(1+x) which comes out to be 2 + x + (3 - 2x - 8x^2 )^0.5 – vikiiii Feb 23 '12 at 8:47
thanks Brian. :) – vikiiii Feb 23 '12 at 8:54
That’s a start, but now you want $g(1+x)$, which is $\big(f(1+x)\big)^{0.5}$, so you want $$\sqrt{2+x+\sqrt{3-2x-8x^2}}\;.$$ Then you need to figure out for which values of $x$ this expression actually makes sense. First worry about the inner square root: you have to choose $x$ so that $3-2x-8x^2\ge 0$, right? Figure out which values of $x$ do that, and then worry about the outer square root. – Brian M. Scott Feb 23 '12 at 8:55
you can put 1+x into f(x) and show it,but you should consider that,when we have square root,domain is all these numbers,for which equation in square root is nnnegative,so in your case,it means that -8*x^2-2*x+3 should be greater or equal to 0,and if we multiply t by -1,this equation must be less then 0 – dato datuashvili Feb 23 '12 at 8:56
up vote 1 down vote accepted

As I wrote in the comments, $$g(1+x)=\sqrt{2+x+\sqrt{3-2x-8x^2}}\;,\tag{1}$$ so the first step is to ensure that $3-2x-8x^2\ge 0$. This quadratic happens to factor: $3-2x-8x^2=(3+4x)(1-2x)$, so it’s equal to $0$ when $x=-\frac34$ and when $x=\frac12$. To see where it’s positive you can check to see where $3+4x$ and $1-2x$ have the same sign, but it’s easier to realize that the graph of $y=3-2x-8x^2$ is a parabola opening down, so it lies above the $x$-axis between the roots $-\frac34$ and $\frac12$. Thus, $3-2x-8x^2\ge 0$ when $-\frac34\le x\le\frac12$. (You must have made a sign error in the calculation in your comment.)

Okay, now we know that the domain is at most the interval $\left[-\frac34,\frac12\right]$, but there might be values of $x$ in that interval that make $$2+x+\sqrt{3-2x-8x^2}\tag{2}$$ negative. Are there?

At $x=-\frac34$, $(2)$ is equal to $2-\frac34=\frac54$, which is certainly not negative. As $x$ increases from $-\frac34$ to $\frac12$, $2+x$ also increases, so it won’t be a problem: it will always be at least $\frac54$. What about $\sqrt{3-2x-8x^2}$? On the interval $\left[-\frac34,\frac12\right]$ it’s always at least $0$, so it won’t be a problem either, and we can be sure that $(2)$ is always at least $\frac54$ when $-\frac34\le x\le\frac12$. In particular, it won’t be negative, so we can take its square root. Thus, the domain of $g(1+x)$ is the whole interval $\left[-\frac34,\frac12\right]$.

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Thanks Brian.Thank you so much – vikiiii Feb 23 '12 at 9:58
M.Scott I need your help.I will be true to you.I have a question and i posted on this site.but later some guy found that question somewhere, so i deleted my question.I read your answer on this question.… .So i thought you can better explain me this question. How many seven-digit numbers divisible by 11 have the sum of their digits equal to 59? I am really sorry as i am asking you question in a comment.I will very thankful to you if you can explain me. – vikiiii Mar 28 '12 at 14:58
@vikiiii: I didn’t have time to think about this today, but I will take a look. – Brian M. Scott Mar 29 '12 at 7:56
M.Scott Thanks so much.Any time whenever you are free. – vikiiii Mar 29 '12 at 8:18
@vikiiii: The maximum possible digit sum is $7\cdot9=63$, and $63-59=4$, so at least $7-4=3$ of the digits must be $9$. We can have $3$ $9$’s and $4$ $8$’s; $4$ $9$’s, $2$ $8$’s, and a $7$; $5$ $9$’s and $2$ $7$’s; $5$ $9$’s, an $8$, and a $6$; of $6$ $9$’s and a $5$; no other combinations add up to $59$. Now you just have to determine which of these five combinations can be arranged to make multiples of $11$. For example with $3$ $9$’s and $4$ $8$’s, we can make $9898988$, $9898889$, $9888989$, and $8898989$, all of which are multiples of $11$. (cont.) – Brian M. Scott Mar 29 '12 at 8:41

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