$$ \left.\begin{array}{l} ext { 14. Given } f(x)=\left\{\begin{array}{cc}2 x^{2}-x & ext { if } x

Question

$$ \left.\begin{array}{l}	ext { 14. Given } f(x)=\left\{\begin{array}{cc}2 x^{2}-x & 	ext { if } x<3 \ 3-x & 	ext { if } x \geq 3\end{array} 	ext {. Find } \lim _{x ightarrow 3} f(x) 	ext {. }ight. \ 	ext { 15. Given } f(x)=\left\{\begin{array}{cc}\frac{1}{x-1} & 	ext { if } x<-1 \ x^{2}+2 x & 	ext { if } x \geq-1\end{array} 	ext {. Find } \lim _{x ightarrow-1} f(x) .ight. \ 	ext { 16. Given } f(x)=\left\{\begin{array}{ll}x^{2}-3 x & 	ext { if } x<2 \ 4+2 x & 	ext { if } x \geq 2\end{array} 	ext { Is } f(x) 	ext { continuous at } 2 	ext { ? }ight. \ 	ext { 17. For what value(s) of } a 	ext { will the function } f(x)=\left\{\begin{array}{c}x^{2}+2 x 	ext { if } x \geq a \ -1\end{array} 	ext { if } x

UpStudy AI · Accepted Answer

Let's solve each of the problems step by step.

### Problem 14
Given the function:
$$
f(x) = \begin{cases}
2x^2 - x & \text{if } x < 3 \
3 - x & \text{if } x \geq 3
\end{cases}
$$
We need to find $$\lim_{x \to 3} f(x)$$.

**Step 1: Calculate the left-hand limit as $$x$$ approaches 3.**
$$
\lim_{x \to 3^-} f(x) = \lim_{x \to 3} (2x^2 - x) = 2(3^2) - 3 = 18 - 3 = 15
$$

**Step 2: Calculate the right-hand limit as $$x$$ approaches 3.**
$$
\lim_{x \to 3^+} f(x) = \lim_{x \to 3} (3 - x) = 3 - 3 = 0
$$

**Step 3: Compare the limits.**
Since $$\lim_{x \to 3^-} f(x) = 15$$ and $$\lim_{x \to 3^+} f(x) = 0$$, the limits are not equal.

Thus, $$\lim_{x \to 3} f(x)$$ does not exist.

### Problem 15
Given the function:
$$
f(x) = \begin{cases}
\frac{1}{x-1} & \text{if } x < -1 \
x^2 + 2x & \text{if } x \geq -1
\end{cases}
$$
We need to find $$\lim_{x \to -1} f(x)$$.

**Step 1: Calculate the left-hand limit as $$x$$ approaches -1.**
$$
\lim_{x \to -1^-} f(x) = \lim_{x \to -1} \frac{1}{x-1} = \frac{1}{-1-1} = \frac{1}{-2} = -\frac{1}{2}
$$

**Step 2: Calculate the right-hand limit as $$x$$ approaches -1.**
$$
\lim_{x \to -1^+} f(x) = \lim_{x \to -1} (x^2 + 2x) = (-1)^2 + 2(-1) = 1 - 2 = -1
$$

**Step 3: Compare the limits.**
Since $$\lim_{x \to -1^-} f(x) = -\frac{1}{2}$$ and $$\lim_{x \to -1^+} f(x) = -1$$, the limits are not equal.

Thus, $$\lim_{x \to -1} f(x)$$ does not exist.

### Problem 16
Given the function:
$$
f(x) = \begin{cases}
x^2 - 3x & \text{if } x < 2 \
4 + 2x & \text{if } x \geq 2
\end{cases}
$$
We need to check if $$f(x)$$ is continuous at $$x = 2$$.

**Step 1: Calculate $$f(2)$$.**
$$
f(2) = 4 + 2(2) = 4 + 4 = 8
$$

**Step 2: Calculate the left-hand limit as $$x$$ approaches 2.**
$$
\lim_{x \to 2^-} f(x) = \lim_{x \to 2} (x^2 - 3x) = 2^2 - 3(2) = 4 - 6 = -2
$$

**Step 3: Calculate the right-hand limit as $$x$$ approaches 2.**
$$
\lim_{x \to 2^+} f(x) = \lim_{x \to 2} (4 + 2x) = 4 + 2(2) = 4 + 4 = 8
$$

**Step 4: Compare the limits and the function value.**
Since $$\lim_{x \to 2^-} f(x) = -2$$, $$\lim_{x \to 2^+} f(x) = 8$$, and $$f(2) = 8$$, the left-hand limit does not equal the right-hand limit.

Thus, $$f(x)$$ is not continuous at $$x = 2$$.

### Problem 17
Given the function:
$$
f(x) = \begin{cases}
x^2 + 2x & \text{if } x \geq a \
-1 & \text{if } x < a
\end{cases}
$$
We need to find the value(s) of $$a$$ for which $$f(x)$$ is continuous at $$x = a$$.

**Step 1: Calculate $$f(a)$$.**
$$
f(a) = a^2 + 2a \quad \text{(since $$a \geq a$$)}
$$

**Step 2: Calculate the left-hand limit as $$x$$ approaches $$a$$.**
$$
\lim_{x \to a^-} f(x) = -1
$$

**Step 3: Calculate the right-hand limit as $$x$$ approaches $$a$$.**
$$
\lim_{x \to a^+} f(x) = a^2 + 2a
$$

**Step 4: Set the limits equal for continuity.**
For $$f(x)$$ to be continuous at $$x = a$$:
$$
-1 = a^2 + 2a
$$
Rearranging gives:
$$
a^2 + 2a + 1 = 0
$$
Factoring:
$$
(a + 1)^2 = 0
$$
Thus, $$a + 1 = 0$$ or $$a = -1$$.

Therefore, the value of $$a$$ for which $$f(x)$$ is continuous at $$x = a$$ is $$a = -1$$.
- Problem 14: $$\lim_{x \to 3} f(x)$$ does not exist. - Problem 15: $$\lim_{x \to -1} f(x)$$ does not exist. - Problem 16: $$f(x)$$ is not continuous at $$x = 2$$. - Problem 17: $$a = -1$$.

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