Solution
- The function has the following properties:
[tex]\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}[/tex]
- 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:
[tex]\begin{gathered} Given, \\ f(x)=ax^2+bx+c \\ \text{ The maximum or minimum value is given at the point where,} \\ x=-\frac{b}{2a} \\ \\ f(x)=5-4x^2 \\ \therefore\text{ Maximum value occurs at:} \\ x=-\frac{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}[/tex]
- 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:
[tex]\text{ The Absolute minimum is }-\infty\text{ and it occurs at }x=-\infty\text{ and }x=\infty[/tex]
- 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:
