28. find the value(s) of the constant k that makes the following function continuous at x = 3. show all…

28. find the value(s) of the constant k that makes the following function continuous at x = 3. show all calculus work!! (5 points)\nf(x)=\begin{cases}e^{(3 - x)}+k^{2}-56, &\text{if }xleq3\\cos(x - 3)+k, &\text{if }x>3end{cases}

28. find the value(s) of the constant k that makes the following function continuous at x = 3. show all calculus work!! (5 points)\nf(x)=\begin{cases}e^{(3 - x)}+k^{2}-56, &\text{if }xleq3\\cos(x - 3)+k, &\text{if }x>3end{cases}

Answer

Explanation:

Step1: Recall continuity condition

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

Step2: Evaluate left - hand limit

When (x\rightarrow3^{-}), 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\rightarrow3^{+}), 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)