Question

1. find the value(s) of the constant k that makes the following function continuous at x = 3. show all calculus work!! (5 points)

f(x)=\begin{cases}e^{(3 - x)}+k^{2}-56, &\text{if }xleq3\cos(x - 3)+k, &\text{if }x>3end{cases}

Explanation:

Step1: Recall continuity condition

For a function to be continuous at \(x = a\), \(\lim_{x
ightarrow a^{-}}f(x)=\lim_{x
ightarrow a^{+}}f(x)\). Here \(a = 3\), so \(\lim_{x
ightarrow 3^{-}}(e^{3 - x}+k^{2}-56)=\lim_{x
ightarrow 3^{+}}(\cos(x - 3)+k)\).

Step2: Evaluate left - hand limit

When \(x
ightarrow3^{-}\), substitute \(x = 3\) into \(e^{3 - x}+k^{2}-56\). We get \(e^{3 - 3}+k^{2}-56=1 + k^{2}-56\).

Step3: Evaluate right - hand limit

When \(x
ightarrow3^{+}\), substitute \(x = 3\) into \(\cos(x - 3)+k\). We get \(\cos(3 - 3)+k=1 + k\).

Step4: Set up the equation

Set \(1 + k^{2}-56=1 + k\). Rearrange it to the quadratic form \(k^{2}-k - 56=0\).

Step5: Solve the quadratic equation

Factor the quadratic equation \(k^{2}-k - 56=(k - 8)(k+7)=0\). Using the zero - product property, if \(ab = 0\), then \(a = 0\) or \(b = 0\). So \(k-8=0\) gives \(k = 8\) and \(k + 7=0\) gives \(k=-7\).

Answer:

\(k = 8\) or \(k=-7\)