Chapter 5.3, Problem 1E

To determine

The local extreme values of the function, and identify any absolute extrema and support your answer graphically.

Answer to Problem 1E

The local minimum value is $-1.25$ at y $x=0.5$ and the absolute minimum is also y $-1.25$ .

Explanation of Solution

Given information :

The given function is $y=x^2-x-1$ .

Calculation:

For the continuous function $f\left(x\right)$ .

At the critical point $c$ :

1. If $f$ changes sign from positive to negative at $c$ ( $f\text{'}>c$ for $x<c$ and $f\text{'}<0$ for $x>c$ ), then $f$ has a local maximum value at $c$ .
2. if $f$ changes sign from negative to positive at $c$ ( $f\text{'}<0$ for $x<c$ and $f\text{'}>c$ for $x>c$ ), then $f$ has a local minimum value at $c$ .
3. if $f$ does not changes its sign at $c$ ( $f\text{'}$ has same sign on both sides of $c$ ), then $f$ has no local extreme value at $c$ .

Now let $f\left(x\right)=x^2-x-1$ .

Find the first derivative of $f\left(x\right)$ :

  $f\text{'}\left(x\right)=2x-1$

Find the critical value by setting $f\text{'}\left(x\right)=0$ :

  $\begin{array}{l}2x-1=0\\ 2x=1\\ x=\frac{1}{2}\\ x=0.5\end{array}$

The critical point partition the x-axis into two intervals as shown below:

  

From the first derivative test the sign of $f\text{'}$ is negative of $x<0.5$ and positive for $x>0.5$ .

Therefore, there is local minimum at $x=0.5$ .

Now, the local minimum value is

  $\begin{array}{l}f\left(0.5\right)={\left(0.5\right)}^2-\left(0.5\right)-1\\ f\left(0.5\right)=0.25-0.5-1\\ f\left(0.5\right)=-1.25\end{array}$

Since, $-1.25$ is the lowest point of the function. Therefore, the absolute minimum is $-1.25$ .

The graph of the function to support the answer is drawn below:

  

Hence,

The local minimum value is $-1.25$ at $x=0.5$ and the absolute minimum is also $-1.25$ .