How do you use the second derivative test to find the relative maxima and minima of the given #f(x)= x^4 - (2x^2) + 3#?

Question

Emma Aldridge · Accepted Answer

A relative maximum is where the first derivative is null and the second derivative is negative.
A relative minimum is where the first derivative is null and the second derivative is positive.

#f(x) = x^4-(2x^2)+3# gives you a curve. The relative maxima and minima of the curve, or the points at which it starts to decrease and then resumes its increase, are what we are looking for. The slope of the curve will be null at these locations. #(df)/dx=4x^3-4x# gives you the variations of this curve's slope. Let's look for the solutions to #4x^3-4x=0# #4x^3-4x=(4x^2-4)x=0# We have three solutions here: #x_0=0;x_+=1;x_(-)=-1# Whether these points are maxima or minima is what we now want to know. A maximum point will have a curve that increases before it is reached and decreases after it is passed. When a point is a minimum, the curve will be falling before it is reached and rising after it is passed. A curve has a positive slope when it is increasing and a negative slope when it is decreasing. Therefore, we wish to determine whether the slope (first derivative) is: at a given point. negative-null-positive #rarr# minimum or positive-null-negative #rarr# maximum We apply the second derivative to achieve this: #(d^2f)/dx^2=12x^2-4# with the #x_0;x_+;x_(-)# points: #12*0^2-4=-4 rarr# the slope is decreasing around 0, therefore we are in a "positive-null-negative" situation, therefore, we have a maximum here. #12*(-1)^2-4=12-4=8 rarr# the slope is increasing around 0, therefore we are in a "negative-null-positive" situation, therefore, we have a minimum here. #12*1^2-4=12-4=8 rarr# minimum.

Aurora Alvarado · Answer

undefined To use the second derivative test to find the relative maxima and minima of the function $ f(x) = x^4 - 2x^2 + 3 $:

1. Find the first derivative of $ f(x) $ and set it equal to zero to find critical points.
2. Find the second derivative of $ f(x) $.
3. Evaluate the second derivative at each critical point.
4. If the second derivative is positive at a critical point, the function has a relative minimum at that point. If the second derivative is negative at a critical point, the function has a relative maximum at that point.

For $ f(x) = x^4 - 2x^2 + 3 $:

1. $ f'(x) = 4x^3 - 4x $
   Set $ f'(x) = 0 $ to find critical points:
   $ 4x^3 - 4x = 0 $
   $ 4x(x^2 - 1) = 0 $
   $ x = 0, \pm 1 $

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

3. Evaluate $ f''(x) $ at the critical points:
   $ f''(0) = -4 $ (negative, so relative maximum)
   $ f''(1) = 8 $ (positive, so relative minimum)
   $ f''(-1) = 8 $ (positive, so relative minimum)

Therefore, the function $ f(x) = x^4 - 2x^2 + 3 $ has relative maxima at $ x = 0 $ and relative minima at $ x = \pm 1 $.