Answer:
d. minimum, 3/2
e. maximum, 0
f. maximum, 8/5
Explanation:
Pre-Solving
We are given the following functions:
d. k(x) = 5x²-15x-50
e. j(x)=12-3x²
f. s(x)=-5x²+16x
We want to find the extremum (i.e. minimum or maximum value) and the x value of the extremum for each of the following functions.
Solving
If the coefficient of x² is positive, then the extremum will be a minimum
If the coefficient of x² is negative, then the extremum will be a maximum.
This means that for k(x)=5x²-15x-50, its extremum will be a minimum, since 5 is positive.
For j(x)=12-3x², this can be rewritten as j(x)= -3x² + 12, so its extremum will be a maximum, since -3 is negative.
For s(x)=-5x²+16x, its extremum will be a maximum, since -5 is negative.
Now, to find the value of x, we can use the formula -b/2a, where b is the coefficient of x, and a is the coefficient of x².
So, for k(x)=5x²-15x-50, it will be 15/2(5) = 15/10 = 3/2.
For j(x)=12-3x² (which is j(x)=-3x²+12), the coefficient in front of x is 0. This means it will be 0/2(-3)=0/-6=0.
For s(x)=-5x²+16x, it will be 16/2(-5)=16/-10=-8/5.