Chapter 5.2, Problem 61STP

To determine

Tocheck:whether the given function has maximum or minimum value, and find the value. Also, state the domain and range of the function.

Answer to Problem 61STP

Maximum value $y=-12$ ; and domain is $\left\{f\left(x\right)|f\left(x\right)<-12\right\}$ .

Explanation of Solution

Given:

The given expression is $f\left(x\right)=-4x^2+8x-16$ .

Calculation:

Consider the following expression,

  $f\left(x\right)=-4x^2+8x-16$

For the above function,

  $a=-4$

As, the graph opens downwards the function has a maximum value.

The coordinate of the vertex will be,

  $\begin{array}{l}x=\frac{-b}{2a}\\ x=\frac{-8}{2\left(-4\right)}\{\text{Substitute these values}}\\ x=1\end{array}$

So, the y- coordinate will be,

Plug the value for x in the above coordinate,

  $\begin{array}{l}y=-4x^2+8x-16\\ y=-4{\left(1\right)}^2+8\left(1\right)-16\\ y=-12\end{array}$

So, the maximum value of the function is $y=-12$ .

The domain of the above function is all real numbers.

The range is all the real numbers less than or equal to $-12$ .

Or, $\left\{f\left(x\right)|f\left(x\right)<-12\right\}$ .

Hence, the given function has maximum value and the value is $y=-12$ .

And the domain belongs to R, and the range is $\left\{f\left(x\right)|f\left(x\right)<-12\right\}$ .