Question

let f(x) = \

$$\begin{cases} 2x^2 - 2 & \\text{if } x < 0 \\\\ k & \\text{if } x = 0 \\\\ 4x - 2 & \\text{if } x > 0 \\end{cases}$$

what value of k will make f(x) continuous at x = 0? answer: blank submit answer next item

Explanation:

Step1: Find left - hand limit as \(x\to0^{-}\)

For \(x < 0\), \(f(x)=2x^{2}-2\). The left - hand limit \(\lim_{x\to0^{-}}f(x)=\lim_{x\to0}(2x^{2}-2)\). Substitute \(x = 0\) into \(2x^{2}-2\), we get \(2(0)^{2}-2=- 2\).

Step2: Find right - hand limit as \(x\to0^{+}\)

For \(x>0\), \(f(x) = 4x - 2\). The right - hand limit \(\lim_{x\to0^{+}}f(x)=\lim_{x\to0}(4x - 2)\). Substitute \(x = 0\) into \(4x-2\), we get \(4(0)-2=-2\).

Step3: Use the definition of continuity at a point

A function \(y = f(x)\) is continuous at \(x = a\) if \(\lim_{x\to a^{-}}f(x)=\lim_{x\to a^{+}}f(x)=f(a)\). At \(x = 0\), we have \(\lim_{x\to0^{-}}f(x)=\lim_{x\to0^{+}}f(x)=-2\). And \(f(0)=k\). For the function to be continuous at \(x = 0\), we need \(k=\lim_{x\to0^{-}}f(x)=\lim_{x\to0^{+}}f(x)\), so \(k=-2\).

Answer:

\(-2\)