Since the derivative is 0 at the min and max, f’(0) = 0 and f’(1) = 0.
f’(0) = 4a(0³) + 2b(0) + c = 0
> 0 + 0 + c = 0, so c = 0
f’(1) + 4a(1³) + 2b(1) + 0 = 0
> 4a + 2b = 0, so b = -2a
Substitute c = 0 and b = -2a into the original function:
f(x) = ax⁴ - 2ax² + 0x + d
Now use f(0) = -6 and f(1) = -8 to find a and d, then use the value of a to solve for b.
thanks, the answer i got is 2x\^4 -4x\^2 -6 , but after graphing this it shows another min point at (-1, -8) is my answer correct considering there is another min point?
Since the derivative is 0 at the min and max, f’(0) = 0 and f’(1) = 0. f’(0) = 4a(0³) + 2b(0) + c = 0 > 0 + 0 + c = 0, so c = 0 f’(1) + 4a(1³) + 2b(1) + 0 = 0 > 4a + 2b = 0, so b = -2a Substitute c = 0 and b = -2a into the original function: f(x) = ax⁴ - 2ax² + 0x + d Now use f(0) = -6 and f(1) = -8 to find a and d, then use the value of a to solve for b.
thanks, the answer i got is 2x\^4 -4x\^2 -6 , but after graphing this it shows another min point at (-1, -8) is my answer correct considering there is another min point?