Question

let (f) be the function defined above. for what values of (b) is (f) continuous at (x = 2)?
a 0.508 only
b 0.647 only
c - 1.282 and 0.508
(f(x)=\begin{cases}e^{bx}&\text{for }xleq2\\1.5x + b&\text{for }x>2end{cases})

Explanation:

Step1: Recall the continuity condition

For a function to be continuous at \(x = a\), \(\lim_{x
ightarrow a^{-}}f(x)=\lim_{x
ightarrow a^{+}}f(x)=f(a)\). Here \(a = 2\), so \(\lim_{x
ightarrow 2^{-}}e^{bx}=\lim_{x
ightarrow 2^{+}}(1.5x + b)\).

Step2: Calculate the left - hand limit

\(\lim_{x
ightarrow 2^{-}}e^{bx}=e^{2b}\).

Step3: Calculate the right - hand limit

\(\lim_{x
ightarrow 2^{+}}(1.5x + b)=1.5\times2 + b=3 + b\).

Step4: Set the left - hand and right - hand limits equal

Set \(e^{2b}=3 + b\). Let \(y = e^{2b}-b - 3\). We can use a numerical method (such as Newton - Raphson method). The derivative of \(y\) with respect to \(b\) is \(y^\prime=2e^{2b}-1\). Starting with an initial guess, we find the root of the equation \(e^{2b}-b - 3 = 0\). By using a graphing utility or a calculator with equation - solving capabilities, we find that \(b\approx0.508\).

Answer:

A. 0.508 only