## Calculus (3rd Edition)

$$a=2 ,\ \ b= 1$$
Given $$f(x)=\left\{\begin{array}{ll} {x^{-1}} & {\text { for } x\frac{1}{2}} \end{array}\right.$$ Since $f(-1)= b-a$ \begin{align*} \lim_{x\to -1^+}f(x) &= \lim_{x\to -1^+}x^{-1}\\ &=-1\\ \lim_{x\to -1^-}f(x) &= \lim_{x\to -1^-}(ax+b)\\ &=b-a \end{align*} Then $f(x)$ is continuous at $x= -1$ when $$b-a=-1\tag{1}$$ Since $f(1/2)= \dfrac{1}{2}a+b$ \begin{align*} \lim_{x\to (1/2)^+}f(x) &= \lim_{x\to (1/2)^+}x^{-1}\\ &=2\\ \lim_{x\to (1/2)^-}f(x) &= \lim_{x\to (1/2)^-}(ax+b)\\ &= \dfrac{1}{2}a+b \end{align*} Then $f(x)$ is continuous at $x= 1/2$ when $$\dfrac{1}{2}a+b=2\tag {2}$$ By solving (1) and (2), we get $$a=2 ,\ \ b= 1$$