A local minimum for f(x) = x^4 - 2x^2 + 1 Is this right? occurs where x = -2 .?

Answer 1

Local Minimum is the point at which the curve gets the minimum value.

For finding this you should first find the derivative of f(x) which is f'(x). Then equate it to zero. The values of x you get there will be the critical points. Now substitute the values of x you got from f'(x)=0 in f(x). You will get a value for each substitution. The lowest value is local minimum.

Here f(x) = x^4 - 2x^2 + 1
then f'(x) = 4x^3 - 4x
Now equate f'(x) to zero.
You will get x = 0, 1, -1.
Now substitute these values in f(x).
f(0) = 1
f(1) = 0
f(-1) = 0

So the local minimum occurs at two points i.e., at 1 and -1.
For further check you can see that f''(x) also satisfies.

Hope you understand.

Answer 2

Start with the first derivative:

f(x) = x^4 - 2x^2 + 1
f'(x) = 4x^3 - 4x = 0
4x(x^2 - 1) = 0

x^2 = 1 can have 2 solutions, -1 or 1
so
x = 0 , -1 or 1

Now do second derivative, if either f''(0), f"(-1) or f"(1), you have a minimum.

f''(x) = 12x^2 - 4
f''(0) = -4, a maximum
f''(-1) = 8, a minimum
f''(1) = 8, a minimum


So f(1) = (1)^4 - 2(1)^2 + 1 = 0
and f(-1) = (-1)^4 - 2(-1)^2 +1 = 0

(-1,0 ) and (1,0) are minimums

Answer 3

No, it's a Maximun at x - 2.

Identify if a point is a maxiumun or minimun when x = -2 of the f(x) = x^4 - 2x² + 1.
The 2nd derivate with a minus sign confirms a min. point.

(1st derivate) dy/dx = 4x^3 - 4x
( 2nd derivate d²y/dx² = 12x² - 4x
When x = -2
d²y/dx² = 12(-2)² - 4(-2)
= 12(4) - (-8)
= 48 + 8
= 56 ( that is, plus 56. → Maximun pt.)

Answer 4

f(-2) = (-2)^4 - (-2)^2 + 1 = 16-4+1=13

To find local minima / maxima, take the derivative of the eq and set to zero:
f'(x) = 0 = 4x^3 - 4x
therefore, when x^3 = x or, x=+1,-1,0
To see if they are local minima, take the second dervative of each:
f''(x) = 12x^2 - 4
f''(+1) = 8 (this is positive, so its a local minima)
f''(-1) = 8 (ditto)
f''(0) = -4 (this is negative, so its a local maxima)

Answer 5

No. It occurs at x=+1,-1 f(x)=(x^2-1)^2. Hence, f(x) is always positive and it attains zero at +1,-1