If a function is to be everywhere differentiable, then it must also be continuous everywhere. This implies that the one sided limits must be equal: \(\displaystyle \tiny {(-1)^2 +b(-1) +4 = b(-1)^2 -2}\).

Next, the one sided limits of the derivative must also be equal. That is, \(\displaystyle \tiny {2a(-1)+b = 2b(-1))}\).

Now we have two missing variables and two equations. Set up a system and solve by elimination or substitution. 

 For a function to be continuous at a point \(\displaystyle a\), \(\displaystyle f(a)\) must exist and \(\displaystyle \lim_{x \rightarrow a}f(x)=f(a)\). 

This is true for all values of \(\displaystyle x\) except \(\displaystyle x=-2\) and \(\displaystyle x=2\).

Therefore, the interval of continuity is \(\displaystyle (-\infty, -2) \cup (-2, 2) \cup \(2, \infty)\).

In order to show this statement is false, we must provide one counterexample. 

For example, let \(\displaystyle f(x) =1, g(x)=x\). Both of these are continuous functions (Their graphes are connected with no "jumps" or "breaks"). Then \(\displaystyle f(x)g(x) = x\), but \(\displaystyle x\) does not have an absolute maximum, or mimimum.

The piecewise function

\(\displaystyle \left\{\begin{matrix} 1;\: x< -5\\0;\:x>-5 \end{matrix}\right.\) 

indicates that \(\displaystyle y\) is one when \(\displaystyle x\) is less than five, and is zero if the variable is greater than five.  At \(\displaystyle x=-5\), there is a hole at the end of the split.  

The limit does not indicate whether we want to find the limit from the left or right, which means that it is necessary to check the limit from the left and right.  From the left to right, the limit approaches 1 as \(\displaystyle x\) approaches negative five.   From the right, the limit approaches zero as \(\displaystyle x\) approaches negative five.

Since the limits do not coincide, the limit does not exist for \(\displaystyle \lim_{x \to -5}\).

For a function to be continuous at a particular point, the limit of the function at that point must be equal to the value of the function at that point. 

 First, notice that 

\(\displaystyle f(3)=\frac{3^2-7(3)+12}{3-3}=\frac{(x-3)(x-4)}{(x-3)}=(x-4)=(3-4)=-1\). 

This means that the function is continuous everywhere.

Next, we must compute the limit. Factor and simplify f(x) to help with the calculation of the limit.

\(\displaystyle f(x)=\frac{x^2-7x+12}{x-3}=\frac{(x-4)(x-3)}{(x-3)}=x-4\)

 \(\displaystyle \lim_{x\to3}(x-4)=-1\)

Thus, the limit as x approaches three exists and is equal to \(\displaystyle -1\), so I and II are true statements. 

The definition of continuity at a point is

\(\displaystyle f(a)=\lim_{x\rightarrow a}f(x)\).

Since \(\displaystyle 9\) is in the interval of the domain of the function, we see \(\displaystyle f(9)=3\)

Also, since the function approaches \(\displaystyle 3\) as \(\displaystyle x\) approaches \(\displaystyle 9\) from the left and the right, we see that

\(\displaystyle \lim_{x\rightarrow 9}\sqrt{x}=3\).

Since 

\(\displaystyle f(9)=\lim_{x\rightarrow9}f(x)=3\)

the function is continuous at \(\displaystyle x=9\).

For a function \(\displaystyle f(x)\) to be continuous at a given point \(\displaystyle (a, f(a))\), it must meet the following two conditions:

1.) The point \(\displaystyle (a, f(a))\) must exist, and

2.) \(\displaystyle \lim_{x\rightarrow a}f(x)=f(a)\).

Examining the above graph, \(\displaystyle f(x)\) is continuous at every possible value of \(\displaystyle x\) except for \(\displaystyle x=0\). Thus, \(\displaystyle f(x)\) is continuous on the interval \(\displaystyle (-\infty,0)\cup(0,\infty)\).

For a function \(\displaystyle f(x)\) to be continuous at a given point \(\displaystyle (a, f(a))\), it must meet the following two conditions:

1.) The point \(\displaystyle (a, f(a))\) must exist, and

2.) \(\displaystyle \lim_{x\rightarrow a}f(x)=f(a)\).

Examining the above graph, \(\displaystyle f(x)\) is continuous at every possible value of \(\displaystyle x\) except for \(\displaystyle x=-1\) and \(\displaystyle x=1\). Thus, \(\displaystyle f(x)\) is continuous on the interval \(\displaystyle (-\infty,-1)\cup(-1,1)\cup(1,\infty)\).

For a function \(\displaystyle f(x)\) to be continuous at a given point \(\displaystyle (a, f(a))\), it must meet the following two conditions:

1.) The point \(\displaystyle (a, f(a))\) must exist, and

2.) \(\displaystyle \lim_{x\rightarrow a}f(x)=f(a)\).

Examining the above graph, \(\displaystyle f(x)\) is continuous at every possible value of \(\displaystyle x\) except for \(\displaystyle x=0\) and \(\displaystyle x=1\). Thus, \(\displaystyle f(x)\) is continuous on the interval \(\displaystyle (-\infty,0)\cup(0,1)\cup(1,\infty)\).

For a function to be continuous at a given point , it must meet the following two conditions:

1.) The point must exist, and

2.) .

Examining the above graph, is continuous at every possible value of except for . Thus, is continuous on the interval .