Question

Let $f(x)=4x^2-3x+1$, $\forall x>\frac{3}{8}$, then number of distinct solutions of the equation $f(x)=f^{-1}(x)$ is:

• A. 0
• B. 1
• C. 2
• D. 4

Answer 1

Answer: (B)

1

Answer 2

To find the number of distinct solutions of the equation $f(x) = f^{-1}(x)$, where $f(x) = 4x^2 - 3x + 1$ for $x > \frac{3}{8}$, we first analyze the function $f(x)$.

1. Determine the monotonicity of $f(x)$:

The function $f(x) = 4x^2 - 3x + 1$ is a quadratic function, representing a parabola opening upwards (since the coefficient of $x^2$ is positive, $a=4>0$). The x-coordinate of the vertex of a parabola $ax^2 + bx + c$ is given by $x = -\frac{b}{2a}$. For $f(x) = 4x^2 - 3x + 1$, we have $a=4$ and $b=-3$. So, the x-coordinate of the vertex is $x = -\frac{-3}{2 \times 4} = \frac{3}{8}$.

The domain of $f(x)$ is given as $x > \frac{3}{8}$. Since the parabola opens upwards and the domain is to the right of its vertex, the function $f(x)$ is strictly increasing in this domain. Alternatively, we can check the derivative: $f'(x) = \frac{d}{dx}(4x^2 - 3x + 1) = 8x - 3$. For $x > \frac{3}{8}$, we have $8x > 8 \times \frac{3}{8} = 3$. So, $8x - 3 > 0$, which means $f'(x) > 0$ for all $x$ in the given domain. This confirms that $f(x)$ is strictly increasing.

2. Relate $f(x) = f^{-1}(x)$ to $f(x) = x$:

For a strictly monotonic function, the solutions to the equation $f(x) = f^{-1}(x)$ are the same as the solutions to the equation $f(x) = x$. This is because if $(a,b)$ is a point on $y=f(x)$, then $(b,a)$ is a point on $y=f^{-1}(x)$. If $f(a)=a$, then $(a,a)$ is on $y=f(x)$, and also $(a,a)$ is on $y=f^{-1}(x)$, meaning $f(a)=f^{-1}(a)=a$. For strictly monotonic functions, there are no other intersection points between $f(x)$ and $f^{-1}(x)$ apart from those on the line $y=x$.

3. Solve the equation $f(x) = x$:

Set $f(x)$ equal to $x$: $4x^2 - 3x + 1 = x$

Rearrange the terms to form a quadratic equation: $4x^2 - 3x - x + 1 = 0$ $4x^2 - 4x + 1 = 0$

4. Find the solutions for $x$:

This quadratic equation can be factored as a perfect square: $(2x)^2 - 2(2x)(1) + 1^2 = 0$ $(2x - 1)^2 = 0$

Taking the square root of both sides: $2x - 1 = 0$ $2x = 1$ $x = \frac{1}{2}$

5. Check if the solution is in the given domain:

The domain of $f(x)$ is $x > \frac{3}{8}$. We need to check if $x = \frac{1}{2}$ satisfies this condition. $\frac{1}{2} = \frac{4}{8}$. Since $\frac{4}{8} > \frac{3}{8}$, the solution $x = \frac{1}{2}$ is valid.

Since there is only one distinct solution ($x = \frac{1}{2}$) that satisfies the given domain, the number of distinct solutions for $f(x) = f^{-1}(x)$ is 1.