g(x) = 5-4x^2

To calculate the maximum values first we need to determine the critical values for g(x) .

Then we will need to determine the first derivative:

==> g'(x) = -8x

Now we know that the critical values are the derivativ'es zeros:

==> -8x = 0

==> x = 0

Then the function has an extreme value when x = 0

The factor of x^2 is negative, then the function has maximum values at x = 0

==> g(0) = 5 - 0 = 5

**Then the maximum values is g(0) = 5**

**Or , the function has a maximum values at the point (0,5)**

To find the maximum of g(x) = 5-4x^2.

We find that x^2 has always a non negative value for all x.

x^2 = 0 when x = 0.

Therefore x ^2 has the least value when x = 0.

Therefore x^2 > = 0 fo all x.

Therefore (x^2)(-1) < = 0 for all x, as we have mutipled by -1, the inequality has reverse.

Therefore -x^2 < =0 for all x.

We now add 5 to both sides:

5 - x^2 < = 0+5 for all values of x.

Therefore g(x) < = 0 is less tha 5 for all values of x.

g(x) = 5, when x = 0.

Therefore g(x) maximum when x= 5. So maximumum of g(x) = g(5) = 5.