Question

Consider the function \(f: \mathbb{R} \rightarrow \mathbb{R}\) defined as \(f(x)=x^{2}+3\). Find \(f([-3,5])\) and \(f^{-1}([12,19]) .\)

Short answer

To summarize, for the function \(f(x) = x^{2} + 3\), we have \(f([-3,5]) = [12,28] and \(f^{-1}([12,19]) = [-4,-3] \cup [3,4]\).

Step-by-step solution

Find the image of the set [-3,5]

First, apply the function \(f(x)\) to the values -3 and 5, the endpoints of the interval. You get \(f(-3)= (-3)^{2}+3=9+3=12\) and \(f(5)=5^{2}+3=25+3=28\). So the interval [-3,5] is mapped to the interval [12,28] under the function \(f(x)\). Therefore, \(f([-3,5]) = [12,28]\).

Find the pre-image of the set [12,19]

To find the pre-image of the interval [12,19] under \(f(x)\), you set up the following inequality for \(f(x)\): \(12 \leq x^{2}+3 \leq 19\). Taking steps to solve this inequality by subtracting 3 from each part, you get \(9 \leq x^{2} \leq 16\), and taking the square root of each part, you get two disjoint intervals since the square root function is applied: the first one from -4 to -3, and the second one from 3 to 4. Therefore, the pre-image of the interval [12,19] under the function \(f(x)\) is \([-4,-3] \cup [3,4]\).

Summarizing the final result

In summary, for the function \(f(x)\) given as \(f(x) = x^{2} + 3\) the image of the set [-3,5] is the interval [12,28] and the pre-image of the set [12,19] is the union of two intervals \([-4,-3] \cup [3,4]\).