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Chapter 1: Problem 96

\(g: R \rightarrow R, g(x)=3+3 \sqrt{x}\) and \(f(g(x))=2-3 \sqrt{x}+x\) then \(\mathrm{f}(\mathrm{x})=\ldots\) (a) \(x^{3}-x^{2}+x-5\) (b) \(x^{3}-9 x^{2}+26 x-22\) (c) \(x^{3}+9 x^{2}-26 x+5\) (d) \(x^{3}+x^{2}-x+5\)

Short Answer

Expert verified
\(f(x) = -x + 5 + \frac{(x-3)^2}{9}\)

Step by step solution

01

Find the inverse of g(x)

To find the inverse of a function, we need to switch the roles of x and g(x), and then solve for the new input \(x\). So for the function \(g(x) = 3 + 3\sqrt{x}\), we will replace \(g(x)\) with \(y\). \(y = 3 + 3\sqrt{x}\) Now, switch the roles of \(x\) and \(y\): \(x = 3 + 3\sqrt{y}\) Now, let's solve for \(y\).
02

Solve for y

To solve for \(y\), we need to isolate it on one side of the equation. Start by subtracting 3: \(x - 3 = 3\sqrt{y}\) Next, divide by 3: \(\frac{x-3}{3} = \sqrt{y}\) Now, square both sides of the equation: \((\frac{x-3}{3})^2 = y\) This simplifies to: \(y = \frac{(x-3)^2}{9}\) Now that we have the inverse function, we can denote it as: \(g^{-1}(x) = \frac{(x-3)^2}{9}\)
03

Substitute the inverse function into f(g(x))

Now we'll replace \(g(x)\) with its inverse function, \(g^{-1}(x)\), in the expression for \(f(g(x))\): \(f(g(x)) = 2 - 3\sqrt{x} + x \Rightarrow f(g^{-1}(x)) = 2 - 3\sqrt{\frac{(x-3)^2}{9}} + \left(\frac{(x-3)^2}{9}\right)\) Since we want to find \(f(x)\), we'll just replace \(g^{-1}(x)\) with \(x\), and we'll have: \(f(x) = 2 - 3\sqrt{\frac{(x-3)^2}{9}} + \frac{(x-3)^2}{9}\)
04

Simplify the expression for f(x)

Let's simplify the expression for \(f(x)\): \(f(x) = 2 - 3\left(\frac{x-3}{3}\right) + \frac{(x-3)^2}{9}\) \(f(x) = 2 - (x-3) + \frac{(x-3)^2}{9}\) \(f(x) = -x + 5 + \frac{(x-3)^2}{9}\) Now let's compare this to the given options (a), (b), (c), and (d).
05

Comparison

Comparing \(f(x) = -x + 5 + \frac{(x-3)^2}{9}\) with the given options, we can see that none of the options match the obtained expression for \(f(x)\). This means that there might be an issue with the given options or the question might have some errors. In this case, the given options do not provide a correct answer to the problem.

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Most popular questions from this chapter

If \(\mathrm{f}(\mathrm{x})=[\mathrm{x} /(\mathrm{x}-1)], \mathrm{x} \neq 1\) then \(\left(\right.\) fofof \(\ldots \mathrm{f}_{(17 \text { times })}(\mathrm{x})\) is equal to \(\ldots \ldots\) (a) \([x /(x-1)]\) (b) \(x\) (c) \([x /(x-1)]^{17}\) (d) \([17 \mathrm{x} /(\mathrm{x}-1)]\)

\(\mathrm{A}=\\{(\mathrm{a}, \mathrm{b}) / \mathrm{b}=2 \mathrm{a}-5\\}\) If \((\mathrm{m}, 5)\) and \((6, \mathrm{n})\) are the member of set \(\mathrm{A}\) then \(\mathrm{m}\) and \(\mathrm{n}\) are respectively (a) 5,7 (b) 7,5 (c) 2,3 (d) 5,3

The range of \(\mathrm{f}(\mathrm{x})=6^{\mathrm{x}}+3^{\mathrm{x}}+6^{-\mathrm{x}}+3^{-\mathrm{x}}+2\) is subset of \(\ldots\) (a) \([-\infty,-2)\) (b) \((-2,3)\) (c) \((6, \infty)\) (d) \([6, \infty)\)

If the function \(\mathrm{f}:[1, \infty) \rightarrow[1, \infty)\) is defined by \(\mathrm{f}(\mathrm{x})=2^{\mathrm{X}(\mathrm{x}-1)}\) then \(\mathrm{f}^{-1}(\mathrm{x})\) is (a) \((1 / 3)^{\mathrm{x}}\) (b) \((1 / 2)\left[1+\sqrt{ \left.\left(1+4 \log _{2} \mathrm{x}\right)\right]}\right.\) (c) \(\left.(1 / 2)\left[1-\sqrt{(} 1+4 \log _{2} \mathrm{x}\right)\right]\) (d) \(2^{[1 /\\{x(x-1)\\}]}\)

The relation \(\mathrm{R}\) defined on the let \(\mathrm{A}=\\{1,2,3,4,5\\}\) by \(R=\left\\{(x, y) /\left|x^{2}-y^{2}\right|<16\right\\} \quad\) is given by (a) \(\\{(1,1),(2,1),(3,1),(4,1),(2,3)\\}\) (b) \(\\{(2,2),(3,2),(4,2),(2,4)\\}\) (c) \(\\{(3,3),(4,3),(5,4),(3,4)\\}\) (d) None of these
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