1)Show all local extrema and inflection points y=-x^4+4x^2-7 The graph to the ri

Question

1)Show all local extrema and inflection points

y=-x^4+4x^2-7

The graph to the right shows the first derivative of a function y = f(x). Select a possible graph f that passes through the point P Use the graph of the function f(x) to locate extrema and identify the intervals where the function is concave up and concave down. Local minimum at x = 3; local maximum at x = -3; concave down on (0, infinity); concave up on (-infinity, 0) Local minimum at x = 3; local maximum at x = -3; concave up on (-infinity, -3) and (3, infinity); concave down on (-3, 3) Local minimum at x = 3; local maximum at x = -3; concave up on (0, infinity); concave down on (-infinity, 0) Local minimum at x = 3; local maximum at x = 3; concave down on (-infinity, -3) and (3, infinity); concave up on (-3, 3)

Explanation / Answer

y=-x^4+4x^2-7 = -(x^4 -4x^2 + 4) + 4 - 7 =

-(x^2 - 2)^2 - 3

We can actually solve this problem without calculating a derivative.

The maximum of y occurs when (x^2 - 2)^2 is a minimum, which occurs when (x^2 - 2) = 0, or

x = plus or minus sqrt(2)

Then, on [-sqrt(2), sqrt(2)] x^2 -2 ranges from 0 at -sqrt(2) and sqrt(2) to -2 at 0. Thus, the square is a maximum at 0.

Thus, -(x^2-2)^2 is a minimum at 0, so the local minimum is at 0.

Thus, we have three local extrema, a minimum at 0 and maxima at plus or minus sqrt(2)

Now, we could have determined this by setting the first derivative to 0.

f'(x) = -4x^3 + 8x = -4x(x^2-2) = -4x(x+sqrt(2))(x-sqrt(2))

Thus, critical points are at - sqrt(2), 0, and sqrt(2)

Take the second derivative. It is -12x^2 + 8

f''(-sqrt(2)) = f''(sqrt(2)) = -12*2 + 8 = -16

This is negative, so we have maxima at plus or minus sqrt(2)

f''(0) = 0 + 8 = 8

This is positive, so we have a minimum at 0.

The graph of the first derivative shows that it is positive at 0 and becomes 0 near plus or minus 5, so we should expect a positive slope at x = 0, with f(x) = 3, and there should be a maximum near 5 and a minimum near -5.

A meets these 4 criteria

For the final graph, clearly, the minimum is at 3 and the maximum is at -3, as all the answers show. As there is a max at -3, it is concave down at -3 and in a neighborhood of -3, and as there is a min at 3, it is concave up there and in a neighborhood. In fact, looking at the graph, it is certainly clear that concave up continues to positive infinity and concave down continues from -3 to -infinity.

Via symmetry, it is reasonable to expect the transition from concave down at -3 to concave up at 3 occurs in the middle, at x = 0.

Thus, C is clearly the answer.

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