Question

If $f(x) = x^3 + 3x^2 - 1$, then number of real solutions of $f(f(x)) = 0$, is

• A. 2
• B. 3
• C. 6
• D. 7

Answer 1

Answer: (D)

7

Answer 2

The problem asks for the number of real solutions of the equation $f(f(x)) = 0$, where $f(x) = x^3 + 3x^2 - 1$.

Let $y = f(x)$. The equation becomes $f(y) = 0$.
So, we first need to find the real roots of $f(y) = 0$, which is $y^3 + 3y^2 - 1 = 0$.
Let $g(y) = y^3 + 3y^2 - 1$. To find the number of real roots of $g(y) = 0$, we analyze the function $g(y)$.
The derivative is $g'(y) = 3y^2 + 6y = 3y(y+2)$.
The critical points are where $g'(y) = 0$, which are $y=0$ and $y=-2$.
The values of $g(y)$ at the critical points are:
$g(0) = 0^3 + 3(0)^2 - 1 = -1$. This is a local minimum.
$g(-2) = (-2)^3 + 3(-2)^2 - 1 = -8 + 12 - 1 = 3$. This is a local maximum.
As $y \to \infty$, $g(y) \to \infty$.
As $y \to -\infty$, $g(y) \to -\infty$.
Since the local maximum value (3) is positive and the local minimum value (-1) is negative, the equation $g(y) = 0$ has three distinct real roots. Let these roots be $y_1, y_2, y_3$.

We can estimate the location of these roots:
$g(-3) = (-3)^3 + 3(-3)^2 - 1 = -27 + 27 - 1 = -1$. Since $g(-3) < 0$ and $g(-2) > 0$, there is a root $y_1 \in (-3, -2)$.
$g(-1) = (-1)^3 + 3(-1)^2 - 1 = -1 + 3 - 1 = 1$. Since $g(-1) > 0$ and $g(0) < 0$, there is a root $y_2 \in (-1, 0)$.
$g(1) = 1^3 + 3(1)^2 - 1 = 1 + 3 - 1 = 3$. Since $g(0) < 0$ and $g(1) > 0$, there is a root $y_3 \in (0, 1)$.

So we have three distinct real roots for $y^3 + 3y^2 - 1 = 0$:
$y_1 \in (-3, -2)$
$y_2 \in (-1, 0)$
$y_3 \in (0, 1)$

Now, for each root $y_i$, we need to solve the equation $f(x) = y_i$, which is $x^3 + 3x^2 - 1 = y_i$, or $x^3 + 3x^2 - (1+y_i) = 0$.
The number of real solutions for $f(x) = k$ depends on the value of $k$. The graph of $f(x) = x^3 + 3x^2 - 1$ has a local maximum at $(-2, f(-2)) = (-2, 3)$ and a local minimum at $(0, f(0)) = (0, -1)$.
A horizontal line $y=k$ intersects the graph of $f(x)$ at:

• 1 real point if $k > 3$ or $k < -1$.
• 3 distinct real points if $-1 < k < 3$.
• 2 real points (one of which is a repeated root) if $k = 3$ or $k = -1$.

Let's consider the number of real solutions for $f(x) = y_i$ for each root $y_i$:

Case 1: $y_1 \in (-3, -2)$.
Since $-3 < y_1 < -2$, we have $y_1 < -1$.
The value $y_1$ is less than the local minimum of $f(x)$.
So, the equation $f(x) = y_1$ has 1 real solution.

Case 2: $y_2 \in (-1, 0)$.
Since $-1 < y_2 < 0$, we have $-1 < y_2 < 3$.
The value $y_2$ is between the local minimum and the local maximum of $f(x)$.
So, the equation $f(x) = y_2$ has 3 distinct real solutions.

Case 3: $y_3 \in (0, 1)$.
Since $0 < y_3 < 1$, we have $-1 < y_3 < 3$.
The value $y_3$ is between the local minimum and the local maximum of $f(x)$.
So, the equation $f(x) = y_3$ has 3 distinct real solutions.

The total number of real solutions for $f(f(x)) = 0$ is the sum of the number of real solutions for $f(x) = y_1$, $f(x) = y_2$, and $f(x) = y_3$.
Total number of solutions = (number of solutions for $f(x) = y_1$) + (number of solutions for $f(x) = y_2$) + (number of solutions for $f(x) = y_3$)
Total number of solutions = 1 + 3 + 3 = 7.

The final answer is 7.