Question

algebra 2a semester online practice
complete this assessment to review what you’ve learned. it will not c
which of the following functions is not one-to-one? (1 point)
h(x) = 3x² - 2
h(x) = -2x + 12
h(x) = 5x³ + 12
h(x) = -7x⁵ - 2

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, 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).

Step2: Analyze \(h(x)=3x^{2}-2\)

The function \(h(x) = 3x^{2}-2\) is a quadratic function. The graph of \(y = ax^{2}+bx + c\) (\(a
eq0\)) is a parabola. For \(y=3x^{2}-2\), \(a = 3>0\), so the parabola opens upwards with vertex at \((0,- 2)\). If we take a horizontal line \(y = k\) where \(k>-2\), it will intersect the graph of \(y = 3x^{2}-2\) at two points \((x_1,y)\) and \((x_2,y)\) where \(x_1=-x_2\) (since \(3x_1^{2}-2=3x_2^{2}-2\) implies \(x_1^{2}=x_2^{2}\) implies \(x_1=\pm x_2\)). For example, if \(y = 1\), then \(3x^{2}-2=1\) gives \(3x^{2}=3\) gives \(x^{2}=1\) gives \(x = 1\) or \(x=-1\). So \(h(x)=3x^{2}-2\) is not one - to - one.

Step3: Analyze \(h(x)=-2x + 12\)

The function \(h(x)=-2x + 12\) is a linear function with slope \(m=-2
eq0\). The graph of a non - vertical linear function \(y=mx + b\) (\(m
eq0\)) is a straight line with a non - zero slope. Any horizontal line will intersect a non - vertical straight line at most once. So by the Horizontal Line Test, \(h(x)=-2x + 12\) is one - to - one.

Step4: Analyze \(h(x)=5x^{3}+12\)

The function \(y = 5x^{3}+12\) is a cubic function. The derivative of \(h(x)\) is \(h^\prime(x)=15x^{2}\geq0\) for all real \(x\), and \(h^\prime(x) = 0\) only when \(x = 0\). The function is strictly increasing (since the derivative is non - negative and zero only at isolated points). A strictly increasing function passes the Horizontal Line Test, so \(h(x)=5x^{3}+12\) is one - to - one.

Step5: Analyze \(h(x)=-7x^{5}-2\)

The function \(y=-7x^{5}-2\) has a derivative \(h^\prime(x)=-35x^{4}\leq0\) for all real \(x\), and \(h^\prime(x) = 0\) only when \(x = 0\). The function is strictly decreasing (since the derivative is non - positive and zero only at isolated points). A strictly decreasing function passes the Horizontal Line Test, so \(h(x)=-7x^{5}-2\) is one - to - one.

Answer:

\(h(x)=3x^{2}-2\) (the first option: \(h(x)=3x^{2}-2\))