Question

show that the function $f(x) = 3(x - 5)^2 + 7$ is not one-to-one.

Explanation:

Step1: Recall the definition of one - to - one function

A function \(y = f(x)\) is one - to - one if for every \(y\) in the range of \(f\), there is exactly one \(x\) in the domain such that \(f(x)=y\). In other words, if \(f(a)=f(b)\) implies \(a = b\) for all \(a,b\) in the domain of \(f\). To show a function is not one - to - one, we can find two different values of \(x\) (say \(x_1\) and \(x_2\) where \(x_1
eq x_2\)) such that \(f(x_1)=f(x_2)\).

Step2: Choose two values of \(x\)

Let's consider the vertex form of a quadratic function. The function \(f(x)=3(x - 5)^2+7\) is a quadratic function in vertex form \(y=a(x - h)^2 + k\), where the vertex is at \((h,k)=(5,7)\) and \(a = 3>0\), so the parabola opens upwards.

Let's take \(x_1=4\) and \(x_2 = 6\).

Step3: Calculate \(f(x_1)\) and \(f(x_2)\)

For \(x_1 = 4\):
\[

$$\begin{align*} f(4)&=3(4 - 5)^2+7\\ &=3(- 1)^2+7\\ &=3\times1 + 7\\ &=3 + 7\\ &=10 \end{align*}$$

\]

For \(x_2=6\):
\[

$$\begin{align*} f(6)&=3(6 - 5)^2+7\\ &=3(1)^2+7\\ &=3\times1+7\\ &=3 + 7\\ &=10 \end{align*}$$

\]

We can see that \(f(4)=f(6) = 10\) and \(4
eq6\).

Answer:

Since \(f(4)=f(6) = 10\) and \(4
eq6\), the function \(f(x)=3(x - 5)^2+7\) is not one - to - one.