Chapter 3.5, Problem 30AYU

In Problems 25-32, use the given functions $f$ and $g$.

(a) $Solvef\left(x\right)=0$.

(b) $Solveg\left(x\right)=0$.

(c) $Solvef\left(x\right)=g\left(x\right)$.

(d) $Solvef\left(x\right)>0$.

(e) $Solveg\left(x\right)\le 0$.

(f) $Solvef\left(x\right)\text{ > }g\left(x\right)$.

(g) $Solvef\left(x\right)\ge \text{ 1}$.

$f\left(x\right)=x^2-\text{ 2}x\text{ + 1}$

$g\left(x\right)=-x^2\text{ + 1}$

To determine

To find:

a. The solution of $f\left(x\right)=0$.

b. The solution of $g\left(x\right)=0$.

c. The solution of $f\left(x\right)=g\left(x\right)$.

d. The solution of $f\left(x\right)>0$.

e. The solution of $g\left(x\right)\le 0$.

f. The solution of $f\left(x\right)>g\left(x\right)$.

g. The solution of $f\left(x\right)\ge 1$.

Answer to Problem 30AYU

Solution:

a. $x=1$

b. $x=1,-1$

c. $x=0,1$

d. $x>1$ or $x<1$

e. $x\le -1$ or $x\ge 1$

f. $x<0$ or $x>1$

g. $x\le 0$ or $x\ge 2$

Explanation of Solution

Given:

The functions are:

$f\left(x\right)=x^2-2x+1$

$g\left(x\right)=-x^2+1$

Calculation:

a. $f\left(x\right)=0$

$x^2-2x+1=0$

${\left(x-1\right)}^2=0$

$x-1=0$

$x=1$

Thus the solution of $f\left(x\right)=0$ is $x=1$.

b. $g\left(x\right)=0$

$-x^2+1=0$

$x^2-1=0$

$\left(x+1\right)\left(x-1\right)=0$

$x=1,-1$

Thus the solutions of $g\left(x\right)=0$ are $x=1,-1$.

c. $f\left(x\right)=g\left(x\right)$

$x^2-2x+1=-x^2+1$

$x^2+x^2-2x+1-1=0$

$2x^2-2x=0$

$2x\left(x-1\right)=0$

$x=0,1$

Thus the solutions of $f\left(x\right)=g\left(x\right)$ are $x=0,1$.

d. $f\left(x\right)>0$

$x^2-2x+1>0$

${\left(x-1\right)}^2>0$

$\left(x-1\right)>0$ or $\left(x-1\right)<0$

$x>1$ or $x<1$

Thus the solutions of $f\left(x\right)>0$ are $x>1$ or $x<1$.

e. $g\left(x\right)\le 0$

$-x^2+1\le 0$

$x^2-1\ge 0$

$\left(x+1\right)\left(x-1\right)\ge 0$

$\left(x+1\right)\le 0$ or $\left(x-1\right)\ge 0$

$x\le -1$ or $x\ge 1$

Thus the solutions of $g\left(x\right)\le 0$ are $x\le -1$ or $x\ge 1$.

f. $f\left(x\right)>g\left(x\right)$

$x^2-2x+1>-x^2+1$

$x^2-2x+1+x^2-1>0$

$2x^2-2x>0$

$2x\left(x-1\right)>0$

$x<0$ or $\left(x-1\right)>0$

$x<0$ or $x>1$

Thus the solutions of $f\left(x\right)>g\left(x\right)$ are $x<0$ or $x>1$.

g. $f\left(x\right)\ge 1$

$x^2-2x+1\ge 1$

$x^2-2x+1-1\ge 0$

$x^2-2x\ge 0$

$x\left(x-2\right)\ge 0$

$x\le 0$ or $\left(x-2\right)\ge 0$

$x\le 0$ or $x\ge 2$

Thus the solutions of $f\left(x\right)\ge 1$ is $x\le 0$ or $x\ge 2$.