Question

does the intermediate value theorem guarantee a value of c in the given interval? if so, find the c - value. if not, explain why not.

1. $f(x)=\frac{x^{2}-x}{x}, f(c)=-1$ on $-2,2$ 41. $f(x)=x^{2}-x, f(c)=-1$ on $-2,2$
2. $f(x)=x^{2}-x, f(c)=5$ on $-2,2$

use the graph of $y = f(x)$ at the right for problems 43 - 52.
find these limits.

1. $lim_{x

ightarrow6}f(x)$ 44. $lim_{x
ightarrow4^{+}}f(x)$ 45. $lim_{x
ightarrow4^{-}}f(x)$

1. $lim_{x

ightarrow4}f(x)$ 47. $lim_{x
ightarrow3}f(x)$ 48. $lim_{x
ightarrow1}f(x)$
determine if $f(x)$ is continuous at these x - values. use correct notation to explain your answer using the definition of continuity from page 34.

1. $x = 1$ 50. $x = 3$ 51. $x = 4$ 52. $x = 6$

Explanation:

Step1: Recall Intermediate - Value Theorem conditions

The Intermediate - Value Theorem states that if a function $y = f(x)$ is continuous on a closed interval $[a,b]$, and $k$ is a number between $f(a)$ and $f(b)$, then there exists at least one number $c$ in the interval $(a,b)$ such that $f(c)=k$.

Step2: Analyze function $f(x)=\frac{x^{2}-x}{x}=x - 1,x

eq0$ on $[-2,2]$
First, simplify $f(x)=\frac{x^{2}-x}{x}$ to $y=x - 1$ for $x
eq0$. The function $f(x)$ is not continuous on $[-2,2]$ because it is not defined at $x = 0$. So, the Intermediate - Value Theorem does not apply.

Step3: Analyze function $f(x)=x^{2}-x$ on $[-2,2]$ for $f(c)=-1$

Let $y=x^{2}-x$. Set $y=-1$, then $x^{2}-x+1 = 0$. Calculate the discriminant $\Delta=b^{2}-4ac$ where $a = 1$, $b=-1$, $c = 1$. So, $\Delta=(-1)^{2}-4\times1\times1=1 - 4=-3<0$. There are no real - valued solutions for $x^{2}-x + 1=0$.

Step4: Analyze function $f(x)=x^{2}-x$ on $[-2,2]$ for $f(c)=5$

Set $x^{2}-x=5$, then $x^{2}-x - 5=0$. Using the quadratic formula $x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$ with $a = 1$, $b=-1$, $c=-5$. We have $x=\frac{1\pm\sqrt{(-1)^{2}-4\times1\times(-5)}}{2\times1}=\frac{1\pm\sqrt{1 + 20}}{2}=\frac{1\pm\sqrt{21}}{2}$. Since $\frac{1-\sqrt{21}}{2}\approx\frac{1 - 4.58}{2}\approx - 1.79\in[-2,2]$ and $\frac{1+\sqrt{21}}{2}\approx\frac{1+4.58}{2}\approx2.79
otin[-2,2]$, $c=\frac{1-\sqrt{21}}{2}$.

Step5: Find limits from the graph

For $\lim_{x
ightarrow6}f(x)$: As $x$ approaches 6 from both sides, the function values approach the $y$ - value at the point on the graph near $x = 6$.
For $\lim_{x
ightarrow4^{+}}f(x)$: As $x$ approaches 4 from the right, we look at the values of the function for $x>4$ getting closer to 4.
For $\lim_{x
ightarrow4^{-}}f(x)$: As $x$ approaches 4 from the left, we look at the values of the function for $x<4$ getting closer to 4.
For $\lim_{x
ightarrow3}f(x)$: Since the left - hand limit and the right - hand limit as $x$ approaches 3 are the same (the function is continuous at $x = 3$ from the graph), we can find the common value.
For $\lim_{x
ightarrow1}f(x)$: Since the left - hand limit and the right - hand limit as $x$ approaches 1 are the same (the function is continuous at $x = 1$ from the graph), we can find the common value.

Step6: Check continuity at given $x$ - values

For $x = 1$: $\lim_{x
ightarrow1}f(x)=f(1)$ (from the graph, the function is continuous at $x = 1$).
For $x = 3$: $\lim_{x
ightarrow3}f(x)=f(3)$ (from the graph, the function is continuous at $x = 3$).
For $x = 4$: The left - hand limit and the right - hand limit are not equal (there is a jump at $x = 4$), so $f(x)$ is not continuous at $x = 4$.
For $x = 6$: The function is not defined at $x = 6$ (open circle), so $f(x)$ is not continuous at $x = 6$.

Answer:

1. The Intermediate - Value Theorem does not apply because $f(x)=\frac{x^{2}-x}{x}$ is not continuous on $[-2,2]$.
2. There is no real - valued $c$ such that $f(c)=-1$ for $f(x)=x^{2}-x$ on $[-2,2]$ since $x^{2}-x + 1=0$ has no real solutions.
3. $c=\frac{1-\sqrt{21}}{2}$

43 - 48: Limits are found by observing the graph as $x$ approaches the given values.

1. $f(x)$ is continuous at $x = 1$.
2. $f(x)$ is continuous at $x = 3$.
3. $f(x)$ is not continuous at $x = 4$.
4. $f(x)$ is not continuous at $x = 6$.