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.