Question

exercises 2.4 continuity
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let ( f(x)=\begin{cases}4x - 1&\text{if }xleq4\\-8x + b&\text{if }x > 4end{cases})
if ( f(x) ) is a function which is continuous everywhere, then we must have
( b=)

Explanation:

Step1: Recall 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 \(a = 4\), so \(\lim_{x
ightarrow4^{-}}f(x)=\lim_{x
ightarrow4^{+}}f(x)\).

Step2: Calculate left - hand limit

For \(x
ightarrow4^{-}\), \(f(x)=4x - 1\). Then \(\lim_{x
ightarrow4^{-}}f(x)=4\times4 - 1=16 - 1 = 15\).

Step3: Calculate right - hand limit

For \(x
ightarrow4^{+}\), \(f(x)=-8x + b\). Then \(\lim_{x
ightarrow4^{+}}f(x)=-8\times4 + b=-32 + b\).

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

Since \(\lim_{x
ightarrow4^{-}}f(x)=\lim_{x
ightarrow4^{+}}f(x)\), we have \(15=-32 + b\).

Step5: Solve for \(b\)

Add 32 to both sides of the equation \(15=-32 + b\), getting \(b=15 + 32=47\).

Answer:

47