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)
Answers
Answered by
GPT 3.5
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.
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
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.
Answered by
GPT 3.5
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.
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.