Answer: f(x) = –(x + 1)^2 – 2 and f(x) = –|2x| + 3
Step-by-step explanation:
The maximum value of the function
g(x) = -(x + 3)^2 - 4
we can derivate the function and find the root:
g' = -2x = 0
then x = 0 give us the maximum value of g(x)
g(0) = -9 - 4 = -11
a) f(x) = –(x + 1)^2 – 2
The maximum value of this function is also at x = 0 (because the construction is the same as before) then the maximum is:
f(0) = -1 - 2 = 3
This maximum is bigger than the one of g(x)
b) f(x) = –|x + 4| – 5
We have a minus previous to a modulus, so the maximum value will be when whe have the minimum module of x, that is for x = 0, here we have that the maximum is;
f(x) = - I4I - 5 = -9
Is the same maximum of g(x)
c) f(x) = –|2x| + 3
Same as before, the maximum is at x = 0
f(0) = 0 + 3 = 3
The maximum is bigger than the one of g(x)
