Chapter 7, Problem 12CR

(a)

To determine

To calculate: The solution of the equation $f(x)=0$ where $f(x)=2x^2+3x+1$ and $g(x)=x^2+3x+2$.

Answer to Problem 12CR

Solution:

The solution set for the equation $f(x)=0$ is $\left\{-1,-\frac{1}{2}\right\}$.

Explanation of Solution

Given information:

The functions $f(x)=2x^2+3x+1$ and $g(x)=x^2+3x+2$.

Formula used:

Quadratic formula for the zeros of the quadratic equation $a{x}^2+bx+c=0$

$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$.

Calculation:

Consider the equation $f(x)=0$.

Solving $f(x)=0$ is equivalent to finding the zeros of the function $f(x)$ using a quadratic formula.

By quadratic formula, the zeros of the quadratic equation $2x^2+3x+1=0$

$x=\frac{-3\pm \sqrt{3^2-4\cdot 2\cdot 1}}{2\cdot 2}=\frac{-3\pm \sqrt{9-8}}{4}=\frac{-3\pm 1}{4}$.

$\Rightarrow x=-\frac{1}{2}\text{or}x=-1$

Thus the solution set for the equation $f(x)=0$ is $\left\{-1,-\frac{1}{2}\right\}$.

(b)

To determine

To calculate: The solution of the equation $f(x)=g(x)$ where $f(x)=2x^2+3x+1$ and $g(x)=x^2+3x+2$.

Answer to Problem 12CR

Solution:

The solution set for the equation $f(x)=g(x)$ is $\left\{-1,1\right\}$.

Explanation of Solution

Given information:

The functions $f(x)=2x^2+3x+1$ and $g(x)=x^2+3x+2$.

Formula used:

Quadratic formula for the zeros of the quadratic equation $a{x}^2+bx+c=0$

$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$.

Calculation:

Consider the equation $f(x)=g(x)$,

Solving $f(x)=g(x)$ is equivalent to find the zeros of the function $f(x)-g(x)$ using quadratic formula.

Now $f(x)-g(x)=2x^2+3x+1-\left(x^2+3x+2\right)$

$\Rightarrow f(x)-g(x)=2x^2+3x+1-x^2-3x-2$

$\Rightarrow f(x)-g(x)=x^2-1$.

By quadratic formula, the zeros of the quadratic equation $x^2-1=0$

$x=\frac{-0\pm \sqrt{0^2-4\cdot 1\cdot \left(-1\right)}}{2\cdot 1}=\frac{\pm \sqrt{4}}{2}=\pm \frac{2}{2}$.

$\Rightarrow x=1\text{or}x=-1$.

Thus the solution set for the equation $f(x)=g(x)$ is $\left\{-1,1\right\}$.

(c)

To determine

The solution of the inequality $f(x)>0$ where $f(x)=2x^2+3x+1$.

Answer to Problem 12CR

Solution:

The inequality $f(x)=2x^2+3x+1>0$ has the solution $\left(-\infty ,-1\right)\cup \left(-\frac{1}{2},\infty \right)$.

Explanation of Solution

Given information:

Polynomials $f(x)=2x^2+3x+1$ and $g(x)=x^2+3x+2$.

Explanation:

Consider the inequality $f(x)>0$,

From part (a), the real zeros of the function $f(x)=2x^2+3x+1$ is $x=-1\text{,}x=-\frac{1}{2}$.

Use the real zeroes to separate the real number line into three intervals as

$(-\infty ,-1),\left(-1,-\frac{1}{2}\right),\left(-\frac{1}{2},\infty \right)$.

To select a test point from each interval and determine whether $f(x)$ is positive or negative.

$\begin{array}{cccc}\text{Interval}& (-\infty ,-1)& \left(-1,-\frac{1}{2}\right)& \left(-\frac{1}{2},\infty \right)\\ \text{Number}\text{chosen}& -4& -\frac{3}{4}& 0\\ \text{Value}\text{of}f& \begin{array}{l}f(-4)=2{\left(-4\right)}^2+3\left(-4\right)+1\\ =21\end{array}& \begin{array}{l}f\left(-\frac{3}{4}\right)=2{\left(-\frac{3}{4}\right)}^2+3\left(-\frac{3}{4}\right)+1\\ =-\frac{1}{8}\end{array}& \begin{array}{l}f(0)=2{\left(0\right)}^2+3\left(0\right)+1\\ =1\end{array}\\ \text{Conclusion}& \text{Positive}& \text{Negative}& \text{Positive}\end{array}$

$f(x)$ is positive for the interval $\left(-\infty ,-1\right)\cup \left(-\frac{1}{2},\infty \right)$.

For the inequality $f(x)>0$, we need to check at theendpoints of the interval also.

For endpoints $x=-1,-\frac{1}{2}$

Endpoint $x=-1$,

$f(-1)=2{\left(-1\right)}^2+3\left(-1\right)+1=2-3+1=0$.

Hence $f(-1)>0$ does not hold for the function $f$.

Endpoint $x=-\frac{1}{2}$,

$f\left(-\frac{1}{2}\right)=2{\left(-\frac{1}{2}\right)}^2+3\left(-\frac{1}{2}\right)+1=0$

Hence $f\left(-\frac{1}{2}\right)>0$ does not hold for the function $f$.

Thus, the endpoints $x=-1,-\frac{1}{2}$ are not included in the solutions.

Therefore, the inequality $2x^2+3x+1>0$ has the solution $\left(-\infty ,-1\right)\cup \left(-\frac{1}{2},\infty \right)$.

(d)

To determine

The solution of the inequality $f(x)\ge g\left(x\right)$, where $f(x)=2x^2+3x+1$ and $g(x)=x^2+3x+2$.

Answer to Problem 12CR

Solution:

The inequality $f(x)\ge g\left(x\right)$ has the solution $\left(-\infty ,-1\right]\cup \left[1,\infty \right)$.

Explanation of Solution

Given information:

The functions $f(x)=2x^2+3x+1$ and $g(x)=x^2+3x+2$.

Explanation:

Solving $f(x)\ge g\left(x\right)$ is equivalent to find an interval where the graph of $f(x)$ is above the graph of $g(x)$.

$\Rightarrow 2x^2+3x+1\ge x^2+3x+2$ as $f(x)=2x^2+3x+1$ and $g(x)=x^2+3x+2$.

Write the inequality so that the polynomial expression on the left side and zero is on the right side.

Subtracting $x^2+3x+2$ from both sides,

$\Rightarrow 2x^2+3x+1-x^2-3x-2\ge 0$

$\Rightarrow x^2-1\ge 0$.

To determine the real zeroes of the function $x^2-1$,

$x^2-1=0$

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

$\Rightarrow x=-1$ or $x=1$

The real zeroes of the function is $x=1\text{,}x=-1$.

Use the real zeroes to separate the real number line into three intervals as

$(-\infty ,-1),(-1,1),(1,\infty )$.

To select a test point from each interval and determine whether $f(x)$ is positive or negative.

$\begin{array}{cccc}\text{Interval}& (-\infty ,-1)& (-1,1)& (1,\infty )\\ \text{Number}\text{chosen}& -4& 0& 4\\ \text{Value}\text{of}{x}^2-1& \begin{array}{l}{\left(-4\right)}^2-1\\ =15\end{array}& \begin{array}{l}{\left(0\right)}^2-1\\ =-1\end{array}& \begin{array}{l}{\left(4\right)}^2-1\\ =15\end{array}\\ \text{Conclusion}& \text{Positive}& \text{Negative}& \text{Positive}\end{array}$

$f(x)-g(x)$ is positive for the interval $\left(-\infty ,-1\right)\cup \left(1,\infty \right)$.

For the inequality $f(x)-g(x)\ge 0$, we need to check at the endpoints of the interval also.

For the endpoints $x=-1,1$

Endpoint $x=-1$,

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

Hence $f(-1)\ge g\left(-1\right)$ holds for the inequality $f(x)\ge g\left(x\right)$.

Endpoint $x=1$,

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

Hence $f(1)\ge g\left(1\right)$ holds for the inequality $f(x)\ge g\left(x\right)$.

Therefore, the endpoints $x=-1$ and $x=1$ are included in the solution.

Thus the inequality $f(x)\ge g\left(x\right)$ has the solution $\left(-\infty ,-1\right]\cup \left[1,\infty \right)$.