What's an example of a short problem where either the first or second derivative test can be applied but the first derivative test is easier? What's an example of a short problem where the second derivative test is easier?
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I don't know how to tell if the first derivative test is easier or the second is easier...how would you figure that out?
An example of a short problem where the first derivative test is easier to apply is finding the local extrema of the function f(x) = x^3 - 3x^2 - 9x + 5.
To apply the first derivative test, we need to find the critical points of the function, which occur where f'(x) = 0 or f'(x) is undefined. Taking the derivative, we get f'(x) = 3x^2 - 6x - 9. Setting this equal to zero and solving for x, we find that x = -1 and x = 3 are the critical points.
By evaluating the sign of f'(x) on either side of these critical points, we can determine the behavior of the function and identify the local extrema. Substituting test values into f'(x), we observe that f'(-2) = 9 and f'(2) = -3.
Thus, at x = -1, f'(x) changes from positive to negative, indicating that a local maximum occurs there. At x = 3, f'(x) changes from negative to positive, indicating that a local minimum occurs there. By evaluating the function at these critical points, we obtain f(-1) = 18 and f(3) = -10.
Therefore, the function has a local maximum at x = -1 and a local minimum at x = 3.
Conversely, an example of a short problem where the second derivative test is easier to apply is finding the local extrema of the function g(x) = x^4 - 4x^2 + 2.
To apply the second derivative test, we first find the critical points by setting g'(x) = 0 and solving for x. Taking the derivative, we get g'(x) = 4x^3 - 8x. Setting this equal to zero yields x = 0 and x = ±√2 as the critical points.
Next, we find the second derivative of g(x), given by g''(x) = 12x^2 - 8. We then evaluate g''(x) at the critical points found earlier. We have g''(0) = -8 and g''(±√2) = 16 - 8 = 8.
According to the second derivative test, if g''(x) > 0 at a critical point, then the function has a local minimum at that point. Conversely, if g''(x) < 0, the function has a local maximum. If g''(x) = 0, the test is inconclusive.
In this case, since g''(0) = -8 < 0, we conclude that g(x) has a local maximum at x = 0. Likewise, since g''(±√2) = 8 > 0, we conclude that g(x) has local minima at x = ±√2.
Therefore, the second derivative test allows us to easily identify the local extrema of g(x).