Question

Continuous Function Find a value for \(a\) so that the function $$f(x)=\left\\{\begin{array}{ll}{x^{2}+x+a,} & {x<1} \\ {x^{3},} & {x \geq 1}\end{array}\right.$$ is continuous.

Short answer

For the function to be continuous, the value of a should be -1.

Step-by-step solution

Formulate the Condition for a Continuous Function

A function \(f(x)\) is continuous at \(x=c\) if \(\lim_{x\to c^-} f(x) = f(c) = \lim_{x\to c^+} f(x)\). Here, as the function changes definitions at \(x=1\), we should apply the condition of continuity at \(x=1\). Therefore, we must ensure that \(\lim_{x\to 1^-}(x^2 + x + a) = f(1) = \lim_{x\to 1^+} x^3\).

Evaluate the Right-hand and Left-hand Limits

In this step, replace \(x\) with 1 in the two function definitions, i.e. \(x^2 + x + a\) is the left-hand limit and \(x^3\) is the right-hand limit. The left-hand limit when \(x=1\) is \(1^2 + 1 + a = 1 + 1 + a = 2 + a\). For the right-hand limit, when \(x=1\), \(f(x) = 1^3 = 1\).

Equate the Two Limits to Solve for \(a\)

The condition for \(f(x)\) to be continuous at \(x=1\) is that the right-hand limit equals the left-hand limit. Equating \(2 + a\) with \(1\) gives an equation for \(a\). Solving this equation gives \(a = 1 - 2 = -1\).