Question

The derivative of a function f is f′(x)=2x(x+5)(x−6) . From this function, which of the following must be true about f? (1 point)

Answer 1

f has a local maximum at x = 0 and local minimums at x=−5, 6 . f has a local maximum at x = 0 and local minimums at x = − 5 , 6 . f has local maximums at x=−5, 6 and a local minimum at x = 0. f has local maximums at x = − 5 , 6 and a local minimum at x = 0. f has a local maximum at x = 2 and local minimums at x=−5, 6 . f has a local maximum at x = 2 and local minimums at x = − 5 , 6 . f has a local minimum at x = 0 and no local maximums. f has a local minimum at x = 0 and no local maximums. f has local minimums at x=−5, 6 and no local maximums.

Answer 2

To find out what must be true about f, we need to integrate the given function f'(x).

Integrating f'(x) gives us f(x) = 2 * integral of (x(x+5)(x-6))dx

Now, we can integrate the function:
f(x) = 2 * integral of (x^3 - x^2 + 5x^2 - 30x)dx
= 2 * integral of (x^3 + 4x^2 - 30x)dx
= 2 * (1/4)x^4 + (4/3)x^3 - (15/2)x^2 + C

Therefore, f(x) = (1/2)x^4 + (8/3)x^3 - (15/2)x^2 + C

From this, we can conclude that f(x) must be a quartic polynomial of the form ax^4 + bx^3 + cx^2 + d, where a = 1/2, b = 8/3, c = -15/2.

Answer 3

From the given derivative function f'(x) = 2x(x+5)(x-6), we can determine the critical points where the derivative is equal to zero or undefined.

Setting f'(x) equal to zero, we have:
2x(x+5)(x-6) = 0

From this equation, we can see that there are three critical points at x = 0, x = -5, and x = 6.

To determine the nature of each critical point, we can use the first derivative test.

For x = -5, plugging in a test value slightly less than -5 (e.g., x = -6), f'(x) will be negative. Plugging in a test value slightly greater than -5 (e.g., x = -4), f'(x) will be positive. Therefore, at x = -5, f has a local minimum.

For x = 0, plugging in a test value slightly less than 0 (e.g., x = -1), f'(x) will be negative. Plugging in a test value slightly greater than 0 (e.g., x = 1), f'(x) will be positive. Therefore, at x = 0, f has a local maximum.

For x = 6, plugging in a test value slightly less than 6 (e.g., x = 5), f'(x) will be positive. Plugging in a test value slightly greater than 6 (e.g., x = 7), f'(x) will be positive. Therefore, at x = 6, f has a local minimum.

So, from the given derivative function, the statement "f has a local maximum at x = 0 and local minimums at x = -5, 6" is true.