Question

Consider the function f(x) = 5 - 4x ^ 2; 2, - 4 <= x <= 2

Answer 1

- The function has the following properties:

`\begin{gathered} f(x)=5-4x^2 \\ \text{ Comparing with the general formula} \\ f(x)=ax^2+bx+c \\ \text{ if }a<0,\text{ the function has a maximum value} \\ \text{ if }a>0,\text{ the function has a minimum value} \end{gathered}`

- The value of "a" is less than 0, therefore, the function has only a maximum value.

- We can find this maximum value by applying the formula given below:

`\begin{gathered} Given, \\ f(x)=ax^2+bx+c \\ \text{ The maximum or minimum value is given at the point where,} \\ x=-(b)/(2a) \\ \\ f(x)=5-4x^2 \\ \therefore\text{ Maximum value occurs at:} \\ x=-(0)/(2(-4))=0 \\ \\ \text{ Put the value of x into the function, to find the maximum value.} \\ f(0)=5-4(0) \\ f(0)=5 \\ \\ \text{ Thus, the MAXIMUM VALUE is 5} \end{gathered}`

- The Absolute maximum value is 5 and it occurs at x = 0

- Since the function is a quadratic, if it has a maximum value, its minimum value is at negative infinity, while if the function has a minimum value, its maximum value is at positive infinity.

- Thus, we have that:

`\text{ The Absolute minimum is }-\infty\text{ and it occurs at }x=-\infty\text{ and }x=\infty`

- Strictly speaking, this implies that there is no Absolute minimum since infinity is not on the number line.

- To understand this further, we can plot the graph as follows: