1. Find all points of inflection of the function f(x)=x^4-6x^3
A. (0, 0)
B. (0, 0), (9/2, -2187/16)
C. (3, -81)
D. (0, 0), 3, -81)
E. none of these
I got D. I found the second derivative and solved for x and plugged values into original function to get points.
2. Find dy/dx for y=sec3x
A. 9sec3xtan3x
B. 3sec3xtan3x
C. 3sec^2(3x)
D. 3tan^2(3x)
E. none of these
I was unsure on this one. I got B. I found the first derivative...
Given that f(x)=-x^2+12x-34 has a relative maximum at x=6, choose the correct statement.
A. f’ is positive on the interval (6, infinity)
B. f’ is positive on the interval (negative infinity, infinity)
C. f’ is negative on the interval (6, infinity)
D. f’ is negative on the interval (negative infinity, 6)
E. none of these
I got C. I found the first derivative and critical number. I used the interval test and first derivative test to determine positive or negative.
Thank you for checking my answers.
all look good to me.
No reason to be unsure on #2. Just the chain rule.
If u = 3x,
y=sec u
y' = secu tanu u'
1. Your answer is correct. To find the inflection points of a function, you need to find the second derivative and solve for x such that f''(x) = 0 or undefined. Then, plug those x-values into the original function to get the corresponding y-values. In this case, the second derivative is f''(x) = 12x^2 - 36x, which equals 0 when x = 0 or x = 3. Plugging these x-values into the original function, we get f(0) = 0 and f(3) = -81, which matches the points given in option D.
2. Your answer is incorrect. To find dy/dx for y = sec(3x), you need to use the chain rule. The derivative of sec(u) is sec(u)tan(u). Therefore, the derivative of y = sec(3x) is dy/dx = 3sec(3x)tan(3x). The correct answer is A.
3. Your answer is incorrect. To determine the intervals where f'(x) is positive or negative to find the relative extrema, you need to use the first derivative test. In this case, the first derivative is f'(x) = -2x + 12. Set f'(x) = 0 to find the critical point, which is x = 6. Then, choose a value to the left of 6 (e.g., x = 5) and a value to the right of 6 (e.g., x = 7) to determine the sign of f'(x) in those intervals. Plugging these x-values into f'(x), we get f'(5) = -2 and f'(7) = -2. Since f'(x) is negative in both the interval (-∞, 6) and (6, ∞), the correct answer is D, f' is negative on the interval (negative infinity, 6).