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Chapter 2: Problem 113

If \(x \geq 0\) \(x^{2}-x+1-2=0\) \(x^{2}-x-1=0\) \(x=1 \pm \sqrt{5} \Rightarrow x=\frac{1+\sqrt{5}}{2}\) If \(x<0\) \(x^{2}-x-2=0\) \((x-2)(x+1)=0\) \(\mathrm{x}=2\) or \(-1\) so, \(x=-1\) No options matches.

Short Answer

Expert verified
Answer: The possible solutions are \(x = \frac{1 + \sqrt{5}}{2}\) for \(x \geq 0\) and \(x = -1\) for \(x < 0\). However, given that there are no matching options for the solutions, it is possible that the answer choices provided were incorrect or the problem might contain an error or typo.

Step by step solution

01

Solve the first quadratic equation

For \(x \geq 0\), the equation given is \(x^2 - x + 1 - 2 = 0\). First, we simplify the equation to get \(x^2 - x - 1 = 0\). Now, we will use the quadratic formula to solve for \(x\), which is: \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) For our equation, a = 1, b = -1, and c = -1. Plugging in these values, we get: \(x = \frac{1 \pm \sqrt{(-1)^2 - 4(1)(-1)}}{2(1)}\)
02

Calculate the first set of solutions

Now, we will calculate the solutions of the first equation: \(x = \frac{1 \pm \sqrt{1 + 4}}{2}\) \(x = \frac{1 \pm \sqrt{5}}{2}\) Since \(x \geq 0\), we choose the positive solution: \(x = \frac{1 + \sqrt{5}}{2}\)
03

Solve the second quadratic equation

For \(x < 0\), the equation given is \(x^2 - x - 2 = 0\). We notice that this equation can be factored as: \((x-2)(x+1) = 0\)
04

Calculate the second set of solutions

Now, we will solve the second equation: \(x - 2 = 0 \Rightarrow x = 2\ (\text{not valid since }x < 0)\) \(x + 1 = 0 \Rightarrow x = -1\) The solution for the second equation is \(x = -1\), which satisfies the condition given, \(x < 0\).
05

Conclusion

We have two possible solutions for the exercise: \(x = \frac{1 + \sqrt{5}}{2}\) for \(x \geq 0\), and \(x = -1\) for \(x < 0\). However, we are given that there are no matching options for the solutions, so it is possible that the answer choices provided were incorrect or the problem might contain an error or typo.

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Most popular questions from this chapter

\(\lim _{x \rightarrow 0} \frac{\sin a x}{b x}=\frac{a}{b}\) \(\lim _{x \rightarrow 0^{\circ}}(a x+1)=1\) $\lim _{x \rightarrow 1}(a x+1)=a+1 \quad \lim _{x \rightarrow 2}\left(c x^{2}-2\right)=4 c-2$ $\lim _{x \rightarrow 1^{\prime}}\left(c x^{2}-2\right)=c-2 \quad \lim _{x \rightarrow 2^{\prime}} \frac{d\left(x^{2}-4\right)}{\sqrt{x}}=0$

$\begin{aligned} f(x) &= \begin{cases}\left|x^{2}-1\right|-1, & x \leq 1 \\\ |2 x-3|-|x-2|, & x>1\end{cases} \\ &=\left\\{\begin{array}{cc}x^{2}-2 & x \leq-1 \\ -x^{2} & -1

$\lim _{x \rightarrow 2} \frac{x-[x]}{x-2}=\lim _{x \rightarrow 2} \frac{\\{x\\}}{\\{x\\}}=1$ So, \(b=1\) $\lim _{x \rightarrow 2} \frac{a\left|x^{2}-x-2\right|}{2+x-x^{2}}=\lim _{x \rightarrow 2} \frac{a\left(2+x-x^{2}\right)}{2+x-x^{2}}$ \(=a\) So, \(a=1\)

\(2 \leq \frac{f(x)}{x^{2}} \leq 3\) \(2 x^{2} \leq f(x) \leq 3 x^{2}\) If \(x=\frac{1}{2}\) $$ \rightarrow \frac{1}{2} \leq f\left(\frac{1}{2}\right) \leq \frac{3}{4} $$ If \(x=1\) If \(x=\frac{1}{3}\) \(\quad \rightarrow 2 \leq f(1) \leq 3\) $\quad \rightarrow \frac{2}{9} \leq f\left(\frac{1}{3}\right) \leq \frac{1}{3}$ $\begin{aligned} \mathrm{f}_{\mathrm{x}}=\frac{1}{4} & \\\$$$ $$$& \rightarrow \frac{1}{8} \leq \mathrm{f}\left(\frac{1}{4}\right) \leq \frac{3}{16} \\ \mathrm{f} x=\frac{1}{8} & \\ & \rightarrow \frac{1}{32} \leq \mathrm{f}\left(\frac{1}{8}\right) \leq \frac{3}{64} \end{aligned}$

$f(x)=\left\\{\begin{array}{lr}2 x-1, & -2 \leq x<0 \\ x+2, & 0 \leq x \leq 2\end{array}\right.$ $g(x)= \begin{cases}{[x],} & -4 \leq x<-2 \\ x+2, & -2 \leq x \leq 4\end{cases}$ \(\lim _{x \rightarrow-2^{2}} f(g(x))=2\) \(\lim _{x \rightarrow-2} f(g(x)=\) Not defined
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