A local minimum for f(x) = x^4 - 2x^2 + 1 occurs _____ Possible solutions below.....?

Question

where x = -2

where x = 0

where x = 1

where x = 2

for no value of x

Answer 1

f'(x)= 4x^3-4x
set f'(x) equal to zero
x=0,1,-1

f''(x)=12x^2-4
positive for x=1 and -1
so x=1 is a local minimum

Answer 2

f'(x) = 4x^3 - 4x
The derivative has 3 roots.
minima and maxima can, but do not have to, occur when f'(x) = 0, or at roots of the derivative.
This particular function is even and symmetrical about 0, so it will have 1 local maximum and 2 local minima.

Answer 3

for f(x) to be minimum

f'(x) = 0

f'(x) = 4x^3 - 4x

f'(x) = 4x(x^2 - 1)

4x = 0 or x^2 - 1 = 0

x = 0 or x = 2 or x = -2

Answer 4

f'(x) = 4x^3-4x
Putting it =0, we get x=0, 1, -1

f''(x)= 12x^2-4

For local minima, f''(x)>0 which is true for both x=1 and -1
Therefore, f is minimum at x=1 and -1 and the minimum value is 0, which is =f(1)=f(-1)

Answer 5

Diff f(x) to f'(x)=4x^3-4x
Let f'(x)=0 to find your stn pts
4x(x-1)(x+1)=0 x=0, -1, +1
Diff f'(x) to f"(x)= 12x^2-4
Sub x=0 into f"(x)=-4<0 (max)
Sub x=-1 into f"(x)=8>0 (mini)
Sub x=+1 into f"(x)=8>0 (mini)
Therefore where x = 1