Question

in exercises 65 - 67, find the value of the constant (a, b, or c) that makes the function continuous.

1. $f(x)=\begin{cases}x^{2}-c& \text{for }x < 5\\4x + 2c& \text{for }xgeq5end{cases}$
2. $f(x)=\begin{cases}2x + 9x^{-1}& \text{for }xleq3\\-4x + c& \text{for }x > 3end{cases}$
3. $f(x)=\begin{cases}x^{-1}& \text{for }x < - 1\ax + b& \text{for }-1leq xleq\frac{1}{2}\\x^{-1}& \text{for }x > \frac{1}{2}end{cases}$
4. define $g(x)=\begin{cases}x + 3& \text{for }x < -1\cx& \text{for }-1leq xleq2\\x + 2& \text{for }x > 2end{cases}$

find a value of c such that g is
a. left - continuous
b. right - continuous
in each case, sketch the graph of g.

Explanation:

Step1: For continuity at x=5, equate left and right limits

Left limit: $5^2 - c = 25 - c$; Right limit: $4(5) + 2c = 20 + 2c$

Step2: Solve for c

$25 - c = 20 + 2c \implies 3c = 5 \implies c = \frac{5}{3}$

Step1: For continuity at x=3, equate left and right limits

Left limit: $2(3) + 9(3)^{-1} = 6 + 3 = 9$; Right limit: $-4(3) + c = -12 + c$

Step2: Solve for c

$9 = -12 + c \implies c = 21$

Step1: Continuity at x=-1: left limit = middle value

Left limit: $(-1)^{-1} = -1$; Middle value: $-a + b$; So $-a + b = -1$

Step2: Continuity at x=1/2: right limit = middle value

Right limit: $(1/2)^{-1} = 2$; Middle value: $\frac{a}{2} + b$; So $\frac{a}{2} + b = 2$

Step3: Solve equations

From Step1: $b = a - 1$. Substitute into Step2: $\frac{a}{2} + a - 1 = 2 \implies a=2, b=1$

Part a: Left-continuous at x=-1 (left limit = function value)

Left limit at x=-1: $-1 + 3 = 2$; Function value: $c(-1) = -c$; So $2 = -c \implies c=-2$

Part b: Right-continuous at x=2 (right limit = function value)

Right limit at x=2: $2 + 2 = 4$; Function value: $c(2) = 2c$; So $4 = 2c \implies c=2$

Answer:

$\frac{5}{3}$