Answer:
Step-by-step explanation:
In order to solve the question, we have to derivate each function.
1) f(x) = x2 +4x -11
Then,
f'(x)= 2x +4
Now, if f'(4)=0 we would have a critical point, which could be a minimum. Let's find out:
f'(4)= 2*4 +4 = 12 ≠ 0 then this function doesn't not have a minimum at (4, -3)
2) f(x) = –2x2 + 16x – 35
Then,
f'(x)= -4x +16
Now, if f'(4)=0 we would have a critical point, which could be a minimum. Let's find out:
f'(4)= -4*4 16 = 4 ≠ 0 then this function have a critical point at (4, -3)
then,
f''(4) =-4 <0 then we have a minimum at (4, -3)
3) f(x) = x2 – 4x + 5
Then,
f'(x)= 2x -4
Now, if f'(4)=0 we would have a critical point, which could be a minimum. Let's find out:
f'(4)= 2*4 -4 = 4 ≠ 0 then this function doesn't not have a minimum at (4, -3)
4) f(x) = 2x2 – 16x + 35
Then,
f'(x)= 4x -16
Now, if f'(4)=0 we would have a critical point, which could be a minimum. Let's find out:
f'(4)= 4*4 -16 = 4 ≠ 0 then this function doesn't not have a minimum at (4, -3)
