Problem 105

Question

Given $f(x)=1+x, \quad-1 \leq x \leq 0$ $=-x, \quad 00$ ix. $f(x)$ is injective function.

Step-by-Step Solution

Verified

Answer

All determined functions are as follows: i. $f_{1}(x)=-x, \quad-1 \leq x<0; \quad=x-1, \quad 0 \leq x<1; \quad=1, \quad x=1$. ii. $f_{2}(x)=-x, \quad-1 \leq x \leq 0; \quad=x-1, \quad 00$. None of the above functions is injective.

1Step 1: Determine $f_{1}(x)=f(g(x))$

To determine $f_{1}(x)=f(g(x))$, plug $g(x)$ into $f(x)$. This results in $f_{1}(x)=-x$ for $-1 \leq x<0$ and $f_{1}(x)=x-1$ for $0 \leq x<1$ and $f_{1}(x)=1$ for $x=1$.

2Step 2: Determine $f_{2}(x)=g(f(x))$

Similarly, to determine $f_{2}(x)=g(f(x))$, plug $f(x)$ into $g(x)$. This gives $f_{2}(x)=-x$ for $-1 \leq x \leq 0$ and $f_{2}(x)=x-1$ for \(0

3Step 3: Determine $f_{3}(x)=f_{1}\left(f_{2}(x)\right)$

To determine $f_{3}(x)=f_{1}\left(f_{2}(x)\right)$, plug $f_{2}(x)$ into $f_{1}(x)$ for each interval. This gives $f_{3}(x)=1$ for $x=-1$, $f_{3}(x)=-x-1$ for \(-1

4Step 4: Determine $f_{4}(x)=f_{2}\left(f_{1}(x)\right)$

To determine $f_{4}(x)=f_{2}\left(f_{1}(x)\right)$, plug $f_{1}(x)$ into $f_{2}(x)$ for each interval. This results in $f_{4}(x)=-x-1$ for $-1 \leq x<0$ and $f_{4}(x)=1-x$ for $0 \leq x \leq 1$.

5Step 5: Determine $f_{5}(x)=f(|x|)$

Computing $f_{5}(x)=f(|x|)$, we recall that $|x|$ equals $x$ if $x \geq 0$ and $-x$ else. Plugging this into $f(x)$ yields $f_{5}(x)=x$ for $-1 \leq x<0$, $f_{5}(x)=1$ for $x=0$, and $f_{5}(x)=-x$ for \(0

6Step 6: Determine $f_{6}(x)=|f(x)|$

By computing $f_{6}(x)=|f(x)|$, we take the absolute value of each part of the piecewise function $f(x)$. The results are $f_{6}(x)=1+x$ for $-1 \leq x \leq 0$ and $f_{6}(x)=x$ for \(0

7Step 7: Determine $f_{7}(x)=\operatorname{sgn}(f(x))$

To figure out $f_{7}(x)=\operatorname{sgn}(f(x))$, take the signum function of each part of $f(x)$. When we apply the signum function, $f_{7}(x)=0$ for $x=-1$, $f_{7}(x)=1$ for \(-1

8Step 8: Determine $f_{8}(x)=f(\operatorname{sgn}(x))$

Lastly, to determine $f_{8}(x)=f(\operatorname{sgn} x)$, substitute the signum of $x$ into $f(x)$. The outcome is $f_{8}(x)=0$ for $x<0$, $f_{8}(x)=1$ for $x=0$, and $f_{8}(x)=-1$ for $x>0$.

9Step 9: Determine Which Functions are Injective

An injective function, also called one-to-one, is a function for which no two different inputs give the same output. By observing each combined function, we can see that the function $f(x)$, consists of multiple intervals where the same output can be achieved by different inputs, hence it is not an injective function. But if also functions like $f_{1}(x)$ are considered, they also show this property, hence also they are not injective. In fact, all combined functions are not injective.

Problem 105

Problem 106

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Problem 105

Given $f(x)=1+x, \quad-1 \leq x \leq 0$ $=-x, \quad 00$ ix. $f(x)$ is injective function.

Given \(\begin{aligned} f(x) &=-1, \quad-2 \leq x \leq 0 \\ &=x-1, \quad 0

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