Chapter 4.7, Problem 39E

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 39E

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

The function has maximum value and the maximum value is $f\left(x\right)=1$ .

Domain of the function is set of all real numbers.

Range of the function is $f\left(x\right)\le 1$ .

Explanation of Solution

Given information:

Given function is

  $f\left(x\right)=-2x^2+4x-1$

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.

  $f\left(x\right)=-2x^2+4x-1$

The above function can be written as

  $\begin{array}{c}f\left(x\right)=-2x^2+4x-1\\ =-2\left(x^2-2x+1\right)+1\\ =-2{\left(x-1\right)}^2+1\end{array}$

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

So, here $a<0$ , the parabola opens down.

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

Maximum value of the function is $k=1$ .

Domain of the function is set of all real numbers.

Range of the function is $f\left(x\right)\le 1$ .

Now, sketch the graph of the function.

  

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

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