(12) 8. use the graph to find the following. write does not exist where appropriate.\na) $lim_{x\rightarrow2}…

(12) 8. use the graph to find the following. write does not exist where appropriate.\na) $lim_{x\rightarrow2}f(x)=$\nb) $lim_{x\rightarrow4^{+}}f(x)=$\nc) $lim_{x\rightarrow4^{-}}f(x)=$\nd) $lim_{x\rightarrow4}f(x)=$\ne) $lim_{x\rightarrow - 5}f(x)=$\nf) $f(3)=$\ng) $f(6)=$\nh) $f(-2.3)=$\ni) list any discontinuities and classify them as removable, infinite or jump.

(12) 8. use the graph to find the following. write does not exist where appropriate.\na) $lim_{x\rightarrow2}f(x)=$\nb) $lim_{x\rightarrow4^{+}}f(x)=$\nc) $lim_{x\rightarrow4^{-}}f(x)=$\nd) $lim_{x\rightarrow4}f(x)=$\ne) $lim_{x\rightarrow - 5}f(x)=$\nf) $f(3)=$\ng) $f(6)=$\nh) $f(-2.3)=$\ni) list any discontinuities and classify them as removable, infinite or jump.

Answer

Explanation:

Step1: Recall limit - definition

The limit of a function as $x$ approaches a value is the $y$ - value the function approaches. For one - sided limits, we look at the behavior of the function as $x$ approaches from the left or right. The derivative at a point is the slope of the tangent line at that point.

Step2: Find $\lim_{x\rightarrow2}f(x)$

As $x$ approaches 2 from both the left and the right, the function values approach - 3. So, $\lim_{x\rightarrow2}f(x)=-3$.

Step3: Find $\lim_{x\rightarrow4^{+}}f(x)$

As $x$ approaches 4 from the right, the function values approach 3. So, $\lim_{x\rightarrow4^{+}}f(x)=3$.

Step4: Find $\lim_{x\rightarrow4^{-}}f(x)$

As $x$ approaches 4 from the left, the function values approach 2. So, $\lim_{x\rightarrow4^{-}}f(x)=2$.

Step5: Find $\lim_{x\rightarrow4}f(x)$

Since $\lim_{x\rightarrow4^{-}}f(x)=2$ and $\lim_{x\rightarrow4^{+}}f(x)=3$, $\lim_{x\rightarrow4}f(x)$ does not exist.

Step6: Find $\lim_{x\rightarrow - 5}f(x)$

As $x$ approaches - 5 from both the left and the right, the function values approach 3. So, $\lim_{x\rightarrow - 5}f(x)=3$.

Step7: Find $f^{\prime}(3)$

The function is a straight - line segment for $x\in[3,4]$. The slope of the line segment from $(3, - 3)$ to $(4,2)$ is $m=\frac{2-( - 3)}{4 - 3}=5$. So, $f^{\prime}(3)=5$.

Step8: Find $f^{\prime}(6)$

For $x\geq4$, the function is a horizontal line $y = 3$. The slope of a horizontal line is 0. So, $f^{\prime}(6)=0$.

Step9: Find $f^{\prime}(-2.3)$

The function is a straight - line segment for $x\in[-5,-2]$. The slope of the line segment from $(-5,3)$ to $(-2,1)$ is $m=\frac{1 - 3}{-2-( - 5)}=-\frac{2}{3}$. So, $f^{\prime}(-2.3)=-\frac{2}{3}$.

Step10: Find discontinuities

At $x = 4$, $\lim_{x\rightarrow4^{-}}f(x)=2$ and $\lim_{x\rightarrow4^{+}}f(x)=3$. This is a jump discontinuity.

Answer:

a) - 3 b) 3 c) 2 d) does not exist e) 3 f) 5 g) 0 h) $-\frac{2}{3}$ i) There is a jump discontinuity at $x = 4$.