1. Check if \(f(c)\) is defined.\[f(-3)=-6\]
2. Check both one-sided limits to check whether the limit exists \[\mathop {\lim}\limits_{x \to 3^+} f(x)=-6\] \[\mathop {\lim}\limits_{x \to 3^-} f(x)=-6\] \[\therefore\] \[\mathop {\lim}\limits_{x \to 3} f(x)=-6\]
3. Check if both value of the function and the two-sided limit equal each other. \[\mathop {\lim}\limits_{x \to 3} f(x)=-6=f(-3)\] \[\therefore\] The function is continuous at \(x=-3\)

First, we need to find whether or not the two function are joined. To do do that we will evaluate each function at \(x=6\)

\[f_{1}\left(6\right)=\sqrt{6-2}=\sqrt{4}=2\]

\[f_{2}\left(6\right)=\frac{1}{6}\left(6\right)^{2}-4=6-4=2\]

Since both function are equal to \(2\) at \(x=6\) they are joined together.

Next we need to find the the interval on which each function is continuous. The square root function can't be negative, so it is continuous on the interval \([2,\infty)\). The quadratic function is defined everywhere; therefore, it is continuous on the interval \((-\infty,\infty)\). The piecewise function will be continuous where those two intervals overlap. In this case that is \([2,\infty)\); therefore, the piecewise function is continuous on the interval \([2,\infty)\).

For the piecewise function to be continuous the each function should be equal to the other at \(x=b\); therefore, we need to find where the two parabolas intersect.

Set the two functions equal to each other and solve for \(x\). \[x^{2}-x=-2x^{2}+4\] \[3x^{2}-x+4=0\] \[\left(x+1\right)\left(3x-4\right)=0\] \[x=-1,\ x=\frac{4}{3}\]  \[\therefore\]

If \(b\) is equal to \(-1\) or \(\frac{4}{3}\), the piecewise function will be continuous.

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When asked to find discontinuities, we need to remember that there are three types: removable discontinuities, jump discontinuities, and vertical asymptotes. The only functions that have jump discontinuities are piecewise functions; therefore, we only need to find the removable discontinuities and the asymptotes of the function.

1. First factor both the numerator and denominator.\[f\left(x\right)=\frac{x+3}{x^{2}-7x+12}=\frac{x+3}{\left(x-3\right)\left(x-4\right)}\]
2. Check whether any of the terms match in both the numerator and denominator. If they do then those are hole or removable discontinuities. In this case, there are none, so there are no removable discontinuities.
3. The remaining terms in the denominator tell us where the vertical asymptotes lie and they occur where the denominator is equal to \(0\). In this case, there are two, one at \(x=3\) and another at \(x=4\).

\[\therefore\]

The function \(f\left(x\right)=\frac{x+3}{x^{2}-7x+12}\) is discontinuous at \(x=3\) and \(x=4\)

1. First factor both the numerator and denominator.\[f\left(x\right)=\frac{x-3}{x^{3}+x^{2}-9x-9}=\frac{x-3}{x^{2}\left(x+1\right)-9\left(x+1\right)}=\] \[\frac{x-3}{\left(x^{2}-9\right)\left(x+1\right)}=\frac{x-3}{\left(x-3\right)\left(x+3\right)\left(x+1\right)}\]
2. Check whether any of the terms match in both the numerator and denominator. If they do then those are hole or removable discontinuities. In this case, the \((x-3)\) is in both the numerator and the denominator; therefore, that term can be reduced to \(1\) and there will be a removable discontinuity at \(x=3\)
3. The remaining terms in the denominator tell us where the vertical asymptotes lie and they occur where the denominator is equal to \(0\). In this case, there are two, one at \(x=3-\) and another at \(x=-4\).

\[\therefore\]

The function \(f\left(x\right)=\frac{x-3}{x^{3}+x^{2}-9x-9}\) is discontinuous at \(x=-3\), \(x=-1\), and \(x=3\).

For the function to be continuous the two functions must be joined at \(3\); therefore, their value must be the equal at \(x=3\)

We can solve this by substituting \(3\) into both functions and then setting them equal to each other to solve for \(c\). We can set them equal to each other because we are looking for a \(c\) in which they are equal to each other at \(3\).

\[f_{1}\left(3\right) = 18x+c\]

f_{2}\left(3\right)=3^{2}+12+9=30

\[30=18+c\]\[c=12\]

\[\therefore\]

When \(c=12\), the piecewise function is continuous.

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To prove that the function has a root in the given interval we need to use the intermediate value theorem; therefore, we need to first evaluate the function at the end points of the interval.

\[f(0)=-2\:and\: f(2)=28\]

Since \(-2

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When asked to prove using the IVT on an assessment, it is best to right your answer in the following style.

\[\]

By the Intermediate Value Theorem, there is a point c in the closed interval \([0,2]\) such that \(f(c) = 0\) because \(f(0)