Question

Let $f(x) = \frac{-x^3 + 15}{7}$ and $\alpha$ be the number of real roots of $f(x) = f^{-1}(x)$, $\beta$ be the number of real roots of $f(x) = x$, then $|\alpha - \beta|$ is

Answer 1

0

Answer 2

The function is given by $f(x) = \frac{-x^3 + 15}{7}$.

We need to find the number of real roots of $f(x) = f^{-1}(x)$, denoted by $\alpha$, and the number of real roots of $f(x) = x$, denoted by $\beta$. We are asked to find $|\alpha - \beta|$.

First, let's find the inverse function $f^{-1}(x)$. Let $y = f(x) = \frac{-x^3 + 15}{7}$. To find the inverse, we swap $x$ and $y$: $x = \frac{-y^3 + 15}{7}$ $7x = -y^3 + 15$ $y^3 = 15 - 7x$ $y = (15 - 7x)^{1/3}$ So, $f^{-1}(x) = (15 - 7x)^{1/3}$.

Now consider the equation $f(x) = x$. $\frac{-x^3 + 15}{7} = x$ $-x^3 + 15 = 7x$ $x^3 + 7x - 15 = 0$ Let $g(x) = x^3 + 7x - 15$. We need to find the number of real roots of $g(x) = 0$. We can analyze the derivative of $g(x)$: $g'(x) = 3x^2 + 7$. Since $x^2 \ge 0$, $3x^2 \ge 0$, so $g'(x) = 3x^2 + 7 \ge 7 > 0$ for all real $x$. Since $g'(x) > 0$ for all real $x$, $g(x)$ is a strictly increasing function. A strictly increasing cubic polynomial has exactly one real root. To confirm the existence of a root, we can check the limits: $\lim_{x \to \infty} g(x) = \lim_{x \to \infty} (x^3 + 7x - 15) = \infty$

$\lim_{x \to -\infty} g(x) = \lim_{x \to -\infty} (x^3 + 7x - 15) = -\infty$

Since $g(x)$ is continuous and goes from $-\infty$ to $\infty$, there must be at least one real root. Since it is strictly increasing, there is exactly one real root. Thus, the number of real roots of $f(x) = x$ is $\beta = 1$.

Now consider the equation $f(x) = f^{-1}(x)$. $\frac{-x^3 + 15}{7} = (15 - 7x)^{1/3}$.

Let's analyze the monotonicity of $f(x)$. $f'(x) = \frac{d}{dx}\left(\frac{-x^3 + 15}{7}\right) = \frac{1}{7}(-3x^2) = -\frac{3x^2}{7}$. Since $x^2 \ge 0$, $f'(x) = -\frac{3x^2}{7} \le 0$ for all real $x$. $f'(x) = 0$ only at $x=0$. For $x \ne 0$, $f'(x) < 0$. This means $f(x)$ is strictly decreasing on $\mathbb{R}$.

For a strictly decreasing function $f$, the equation $f(x) = f^{-1}(x)$ is equivalent to $f(x) = x$. Thus, the number of real roots of $f(x) = f^{-1}(x)$ is equal to the number of real roots of $f(x) = x$. Therefore, $\alpha = 1$.

We need to find $|\alpha - \beta|$. $|\alpha - \beta| = |1 - 1| = 0$.