Chapter 4.3, Problem 1E

To determine

The local extreme values and any absolute extremum of the given function.

Answer to Problem 1E

It has been determined that $y\left(\frac{1}{2}\right)=-\frac{5}{4}$ is the only local extreme value of the given function. More specifically, it is a local minimum value. Moreover, it is also the absolute minima of the given function.

Explanation of Solution

Given:

The function, $y=x^2-x-1$ .

Concept used:

The critical points of a function $y\left(x\right)$ are the points $x=c$ such that $y^{\prime }\left(c\right)=0$ .

If $y^{\prime }\left(x\right)<0$ for $x<c$ and $y^{\prime }\left(x\right)>0$ for $x>c$ , then the function $y\left(x\right)$ has a local minima at $x=c$ and vice versa.

The local minima (or maxima) where the function attains the least (or greatest) value is called the absolute minima (or maxima) of the function.

Calculation:

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

Differentiating,

  $y^{\prime }\left(x\right)=2x-1$

Equating the first derivative to zero to obtain the critical points,

  $2x-1=0$

Solving,

  $x=\frac{1}{2}$

So, the only critical point of the given function is $x=\frac{1}{2}$ .

Put $x=0$ in $y^{\prime }\left(x\right)=2x-1$ to get,

  $y^{\prime }\left(0\right)=-1<0$

Put $x=1$ in $y^{\prime }\left(x\right)=2x-1$ to get,

  $y^{\prime }\left(1\right)=1>0$

This implies that $y^{\prime }\left(x\right)<0$ for $x<\frac{1}{2}$ and $y^{\prime }\left(x\right)>0$ for $x>\frac{1}{2}$ , which further implies that $x=\frac{1}{2}$ is a local minima of the given function.

Put $x=\frac{1}{2}$ in $y\left(x\right)=x^2-x-1$ to get,

  $\begin{array}{c}y\left(\frac{1}{2}\right)={\left(\frac{1}{2}\right)}^2-\left(\frac{1}{2}\right)-1\\ =\frac{1}{4}-\frac{1}{2}-1\\ =\frac{1-2-4}{4}\\ =-\frac{5}{4}\end{array}$

So, $y\left(\frac{1}{2}\right)=-\frac{5}{4}$ is the only local extreme value of the given function. More specifically, it is a local minimum value.

Since this is the only critical point, it follows that this is the only local minima.

Thus, by default, it is the absolute minima of the function.

Conclusion:

It has been determined that $y\left(\frac{1}{2}\right)=-\frac{5}{4}$ is the only local extreme value of the given function. More specifically, it is a local minimum value. Moreover, it is also the absolute minima of the given function.