Question

which of the following functions is one-to-one? (1 point)

• ( f(x) = -6x^3 + 7 )
• ( f(x) = -6x^2 + 7 )
• ( f(x) = x^4 )
• ( f(x) = 2x^2 - 4x + 7 )

Explanation:

Step1: Recall one - to - one function definition

A function \(y = f(x)\) is one - to - one (injective) if for every \(y\) in the range of \(f\), there is exactly one \(x\) in the domain such that \(f(x)=y\). Geometrically, a function is one - to - one if no horizontal line intersects its graph more than once (Horizontal Line Test). Also, for a function \(y = f(x)\), if it is strictly increasing or strictly decreasing over its domain, then it is one - to - one.

Step2: Analyze \(f(x)=- 6x^{3}+7\)

The derivative of \(f(x)=-6x^{3}+7\) using the power rule \((x^{n})^\prime=nx^{n - 1}\) is \(f^\prime(x)=-18x^{2}\). The square of a real number \(x^{2}\geq0\), so \(- 18x^{2}\leq0\). And \(f^\prime(x) = 0\) only when \(x = 0\). For \(x
eq0\), \(f^\prime(x)<0\). So the function \(f(x)=-6x^{3}+7\) is strictly decreasing on \((-\infty,0)\) and \((0,\infty)\) (and since the derivative at \(x = 0\) is \(0\) but the function is still decreasing around \(x = 0\) in the sense that for \(x_1<0

Step3: Analyze \(f(x)=-6x^{2}+7\)

This is a quadratic function with the form \(y = ax^{2}+bx + c\) where \(a=-6\), \(b = 0\), \(c = 7\). The graph of a quadratic function \(y=ax^{2}+bx + c\) (\(a
eq0\)) is a parabola. For \(y=-6x^{2}+7\), the parabola opens downwards (since \(a=-6<0\)) with vertex at \((0,7)\). If we take \(y = 1\), then \(-6x^{2}+7 = 1\Rightarrow-6x^{2}=-6\Rightarrow x^{2}=1\Rightarrow x=\pm1\). So the horizontal line \(y = 1\) intersects the graph at \(x = 1\) and \(x=-1\), so it fails the Horizontal Line Test and is not one - to - one.

Step4: Analyze \(f(x)=x^{4}\)

The derivative of \(f(x)=x^{4}\) is \(f^\prime(x) = 4x^{3}\). For \(x<0\), \(f^\prime(x)<0\) (function is decreasing), for \(x>0\), \(f^\prime(x)>0\) (function is increasing). The graph of \(y = x^{4}\) is a "U - shaped" curve (similar to a parabola but flatter near the origin) opening upwards with vertex at \((0,0)\). If we take \(y = 16\), then \(x^{4}=16\Rightarrow x=\pm2\) (since \(2^{4}=16\) and \((-2)^{4}=16\)). So the horizontal line \(y = 16\) intersects the graph at \(x = 2\) and \(x=-2\), so it fails the Horizontal Line Test and is not one - to - one.

Step5: Analyze \(f(x)=2x^{2}-4x + 7\)

First, we can rewrite the quadratic function in vertex form. \(f(x)=2(x^{2}-2x)+7=2(x^{2}-2x + 1-1)+7=2((x - 1)^{2}-1)+7=2(x - 1)^{2}-2 + 7=2(x - 1)^{2}+5\). The graph of this function is a parabola opening upwards (since \(a = 2>0\)) with vertex at \((1,5)\). For \(y=7\), we have \(2(x - 1)^{2}+5=7\Rightarrow2(x - 1)^{2}=2\Rightarrow(x - 1)^{2}=1\Rightarrow x-1=\pm1\Rightarrow x=1 + 1=2\) or \(x=1-1=0\). So the horizontal line \(y = 7\) intersects the graph at \(x = 0\) and \(x = 2\), so it fails the Horizontal Line Test and is not one - to - one.

Answer:

\(f(x)=-6x^{3}+7\) (the first option: \(f(x)=-6x^{3}+7\))