exercises 2.4 continuity\nscore: 5/14 answered: 6/14\n× question 6\ntextbook videos +\nscore on last try: 0…

exercises 2.4 continuity\nscore: 5/14 answered: 6/14\n× question 6\ntextbook videos +\nscore on last try: 0 of 1 pts. see details for more.\n> next question get a similar question you can retry this ques\nlet ( f(x)=\begin{cases}4x - 1&\text{if }xleq4\\-8x + b&\text{if }x > 4end{cases})\nif ( f(x) ) is a function which is continuous everywhere, then we must have\n( b=)

Answer

Explanation:

Step1: Recall continuity condition

For a function to be continuous at (x = a), (\lim_{x\rightarrow a^{-}}f(x)=\lim_{x\rightarrow a^{+}}f(x)=f(a)). Here (a = 4), so (\lim_{x\rightarrow4^{-}}f(x)=\lim_{x\rightarrow4^{+}}f(x)).

Step2: Calculate left - hand limit

For (x\rightarrow4^{-}), (f(x)=4x - 1). Then (\lim_{x\rightarrow4^{-}}f(x)=4\times4 - 1=16 - 1 = 15).

Step3: Calculate right - hand limit

For (x\rightarrow4^{+}), (f(x)=-8x + b). Then (\lim_{x\rightarrow4^{+}}f(x)=-8\times4 + b=-32 + b).

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

Since (\lim_{x\rightarrow4^{-}}f(x)=\lim_{x\rightarrow4^{+}}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).