Sovi.AI - AI Math Tutor

Scan to solve math questions

QUESTION IMAGE

determine whether the function is one-to-one f(x) = 3x³ - 5 is the func…

Question

determine whether the function is one-to-one
f(x) = 3x³ - 5
is the function one-to-one? select the correct choice below and, if necessary, fill in the answer boxes to complete your choice.
○ a. no, because the f(x) value 2 corresponds to two x-values □ and □
○ b. yes, because each x-value corresponds to only one f(x) value, and each f(x) value corresponds to only one x-value

Explanation:

Step1: Recall one - to - one function definition

A function \(y = f(x)\) is one - to - one if whenever \(f(a)=f(b)\), then \(a = b\). Also, we can use the horizontal line test: if no horizontal line intersects the graph of the function more than once, the function is one - to - one. For the function \(f(x)=3x^{3}-5\), we can analyze its monotonicity. The derivative of \(f(x)\) is \(f^\prime(x)=9x^{2}\geq0\) for all real \(x\), and \(f^\prime(x) = 0\) only when \(x = 0\). So the function is strictly increasing (since it's increasing everywhere and only has a horizontal tangent at \(x = 0\), not a flat region). A strictly increasing function is one - to - one because if \(x_1

Step2: Analyze the options

  • Option A: Let's check if \(f(x)=2\). Set \(3x^{3}-5 = 2\), then \(3x^{3}=7\), \(x^{3}=\frac{7}{3}\), and \(x=\sqrt[3]{\frac{7}{3}}\). There is only one real solution for \(x\) when \(f(x) = 2\) (since the cube function is one - to - one). So option A is incorrect.
  • Option B: Since the function \(f(x)=3x^{3}-5\) is a cubic function with a positive leading coefficient and its derivative \(f^\prime(x)=9x^{2}\) shows that it is non - decreasing (and strictly increasing except at \(x = 0\)), each \(x\) - value gives exactly one \(f(x)\) - value (by the definition of a function), and each \(f(x)\) - value comes from exactly one \(x\) - value (because the function is strictly increasing, so it passes the horizontal line test). So the function is one - to - one.

Answer:

B. Yes, because each \(x\) - value corresponds to only one \(f(x)\) value, and each \(f(x)\) value corresponds to only one \(x\) - value