Suppose f(x) = begin{cases} a + bx, & x

Question

(A) The function is continuous at $x = 1$ if $\lim_{x 	o 1^-} f(x) = \lim_{x 	o 1^+} f(x) = f(1)$.First,calculate the left-hand limit:$\lim_{x 	o 1

Vedclass · Accepted Answer

$a = 0, b = 4$

Vedclass · Answer

(A) The function is continuous at $x = 1$ if $\lim_{x 	o 1^-} f(x) = \lim_{x 	o 1^+} f(x) = f(1)$.First,calculate the left-hand limit:$\lim_{x 	o 1^-} f(x) = \lim_{x 	o 1} (a + bx) = a + b$.Next,calculate the right-hand limit:$\lim_{x 	o 1^+} f(x) = \lim_{x 	o 1} (b - ax) = b - a$.Given $f(1) = 4$,we have the system of equations:$a + b = 4$$b - a = 4$Adding the two equations: $(a + b) + (b - a) = 4 + 4$ $\Rightarrow 2b = 8$ $\Rightarrow b = 4$.Substituting $b = 4$ into $a + b = 4$: $a + 4 = 4 \Rightarrow a = 0$.Therefore,the values are $a = 0$ and $b = 4$.

Suppose $f(x) = \begin{cases} a + bx, & x < 1 \\ 4, & x = 1 \\ b - ax, & x > 1 \end{cases}$ and if $\lim_{x \to 1} f(x) = f(1)$,what are the possible values of $a$ and $b$?

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