consider the function; f(x)=x^4-3x^2-1
a) Find all the point where f'(x)=0
b) Use the second derivative test to classify the stationary points?
where do you get stuck?
f' = 4x^3 - 6x
= 2x(2x^2 - 3)
f'' = 12x^2 - 6
= 6(2x^2 - 1)
Thank you so much.
and what about the stationary points ?
To find the points where f'(x) = 0, we need to find the critical points of the function f(x).
a) To find the critical points, we need to find the derivative of f(x), which is f'(x), and then solve the equation f'(x) = 0.
Let's find the derivative of f(x):
f(x) = x^4 - 3x^2 - 1
Take the derivative of each term:
f'(x) = d/dx(x^4) - d/dx(3x^2) - d/dx(1)
Using the power rule of differentiation:
f'(x) = 4x^3 - 6x
Now, to find the points where f'(x) = 0, we solve the equation:
4x^3 - 6x = 0
Factor out the common term:
2x(2x^2 - 3) = 0
Set each factor equal to zero:
2x = 0 and 2x^2 - 3 = 0
Solving the first equation:
2x = 0
x = 0
Solving the second equation using the quadratic formula:
2x^2 - 3 = 0
x^2 = 3/2
x = ±sqrt(3/2)
x ≈ ±1.225
Therefore, the points where f'(x) = 0 are x = 0, x ≈ 1.225, and x ≈ -1.225.
b) To classify the stationary points, we need to use the second derivative test. We find the second derivative, f''(x), and evaluate it at the critical points.
To find the second derivative, we differentiate f'(x):
f'(x) = 4x^3 - 6x
Differentiating f'(x), we get:
f''(x) = d/dx(4x^3) - d/dx(6x)
Using the power rule again:
f''(x) = 12x^2 - 6
Now, let's evaluate f''(x) at the critical points.
At x = 0:
f''(0) = 12(0)^2 - 6
f''(0) = -6
At x ≈ 1.225:
f''(1.225) = 12(1.225)^2 - 6
f''(1.225) ≈ 10.755
At x ≈ -1.225:
f''(-1.225) = 12(-1.225)^2 - 6
f''(-1.225) ≈ 10.755
Using the second derivative test:
1. f''(x) < 0 at x = 0. This means that the point (0, f(0)) is a local maximum.
2. f''(x) > 0 at x ≈ 1.225 and x ≈ -1.225. This means that the points (1.225, f(1.225)) and (-1.225, f(-1.225)) are local minimums.
So, the stationary points of the function f(x) = x^4 - 3x^2 - 1 are:
- A local maximum at (0, f(0))
- Two local minimums at (1.225, f(1.225)) and (-1.225, f(-1.225))