If the function [math: f(x)=x3+4x2+4x] f(x)=x^3+4x^2+4x has a stationary point at (-2,0), the other stationary point is:

Question

BlackTom AI · Accepted Answer

We start by identifying the stationary points of f(x). A stationary point occurs where f'(x) = 0.

Compute f'(x): f(x) = x^3 + 4x^2 + 4x, so f'(x) = 3x^2 + 8x + 4.

Solve 3x^2 + 8x + 4 = 0. The discriminant is Δ = 8^2 - 4·3·4 = 64 - 48 = 16, so the roots are x = [-8 ± √16] / (2·3) = [-8 ± 4] / 6.
This gives x = (-8 + 4)/6 = -4/6 = -2/3 and x = (-8 - 4)/6 = -12/6 = -2.

We already know one stationary point is at x = -2 with y = f(-2) = (-2)^3 + 4(-2)^2 + 4(-2) = -8 + 16 - 8 = 0, i.e., (-2, 0).

Now find the other stationary point by evaluating f at x = -2/3:
f(-2/3) = (-2/3)^3 + 4(-2/3)^2 + 4(-2/3) = (-8/27) + 4(4/9) + (-8/3).
Compute each term with a common denominator 27: (-8/27) + (16/9) + (-8/3) = (-8/27) + (48/27) + (-72/27) = (-32/27).
Thus the second stationary point is at x = -2/3 and y = -32/27, i.e., (-2/3, -32/27).

Now match with the given options:
- Option a: (13,12227) with (1/3, 1 22/27) is clearly not matching the actual coordinates; both the x and y values are far off and the form is inconsistent with the cubic’s stationary points.
- Option b: (2, 32) is not correct because the x-value for a stationary point of f'(x) = 0 is not 2 (we found the critical x-values are -2 and -2/3), and the y-value 32 does not align with f(2).
- Option c: (−2/3, −32/27) is exactly the computed second stationary point, written in a mixed-number form for the y-coordinate as −1 5/27, which equals −32/27.
- Option d: (−2/3, 4 20/27) has the correct x-coordinate but an incorrect y-coordinate, since f(-2/3) evaluates to −32/27, not + (4 20/27).

Therefore, option c correctly matches the second stationary point, while the other options contain incorrect x-values, incorrect y-values, or both, due to arithmetic errors or misinterpretation of the cubic’s derivative roots.

If the function [math: f(x)=x3+4x2+4x] f(x)=x^3+4x^2+4x has a stationary point at (-2,0), the other stationary point is:

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