Question

In Exercises \(11-18,\) use the function \(f\) defined and graphed below to answer the questions. $$f(x)=\left\\{\begin{array}{ll}{x^{2}-1,} & {-1 \leq x<0} \\\ {2 x,} & {0 < x < 1} \\ {1,} & {x=1} \\ {-2 x+4,} & {1 < x < 2} \\ {0,} & {2 < x < 3}\end{array}\right.$$ What value should be assigned to \(f(2)\) to make the extended function continuous at \(x=2 ?\)

Short answer

To make the function \(f(x)\) continuous at \(x = 2\), the value \(f(2) = 0\) should be assigned.

Step-by-step solution

Calculate the left-hand limit

The left-hand limit is equivalent to the limit of the function as we approach the given point from the left. In terms of x, it is \(\lim_{{x \to 2^{-}}}f(x)\). Since for \(x < 2\), the functional form of \(f\) is given by the equation \(-2x + 4\), \(\lim_{{x \to 2^{-}}}{-2x + 4} = 0\). Therefore, the left-hand limit equals \(0\).

Calculate the right-hand limit

The right-hand limit is equivalent to the limit of the function as we approach the given point from the right. Represented mathematically, it is \(\lim_{{x \to 2^{+}}}f(x)\). For \(x \geq 2\), the functional form of \(f\) is given by the equation \(0\), therefore, \(\lim_{{x \to 2^{+}}}{0} = 0\). Thus, the right-hand limit also equals 0.

Make the function continuous at \(x = 2\)

In order for the function to be continuous at a point, the left-hand limit, right-hand limit, and the function value at that point should be equal. From Step 1 and Step 2, we found that both the left-hand limit and the right-hand limit equal \(0\). Hence, to make the function continuous at \(x = 2\), we need to assign the value of \(f(2)\) as \(0\) as well.