Chapter 4.7, Problem 38E

To determine

To find: the maximum and minimum value of the given function. Also describe the domain and range of the function and the interval where the graph is increasing and decreasing.

Answer to Problem 38E

The function is increasing on $\left(1,\infty \right)$ and decreasing on $\left(-\infty ,1\right)$ .

The function has minimum value and the minimum value is $h\left(x\right)=-3$ .

Domain of the function is set of all real numbers.

Range of the function is $h\left(x\right)\ge -3$ .

Explanation of Solution

Given information:

Given function is

  $h\left(x\right)=\frac{1}{2}{\left(x-1\right)}^2-3$

Concept used:

The function $g\left(x\right)=a{\left(x+h\right)}^2+k$ has minimum value if $a>0$ and the parabola opens up.

The function $g\left(x\right)=a{\left(x+h\right)}^2+k$ has maximum value if $a<0$ and the parabola opens down.

The minimum or maximum value occurs at $x=h$ and the minimum or maximum value is $g\left(x\right)=k$ .

Where $\left(h,k\right)$ is the vertex of parabola.

Domain of a function $y=f\left(x\right)$ is all $x$ values for which $y=f\left(x\right)$ is defined.

Range of a function $y=f\left(x\right)$ is all $y$ values for which $y=f\left(x\right)$ is defined.

Calculation:

Consider the given function.

  $h\left(x\right)=\frac{1}{2}{\left(x-1\right)}^2-3$

The function $h\left(x\right)=\frac{1}{2}{\left(x-1\right)}^2-3$ is of the form $g\left(x\right)=a{\left(x+h\right)}^2+k$ where $a=\frac{1}{2},h=-1,k=-3$ .

So, here $a>0$ , the parabola opens up.

The function has minimum value at $x=h=-1$ .

Minimum value of the function is $k=-3$ .

Domain of the function is set of all real numbers.

Range of the function is $h\left(x\right)\ge -3$ .

Now, sketch the graph of the function.

  

The function is increasing to the right of $x=1$ that is on $\left(1,\infty \right)$ .

The function is decreasing to the left of $x=1$ that is on $\left(-\infty ,1\right)$ .