Answer to Question #203447 in Calculus for jay

Question

For what value of b is the function 𝑓(𝑥) = { 𝑥 2 − 1 𝑥 < 3 2𝑏𝑥 𝑥 ≥ 3 continuous everywhere

Answer 1

$$\lim\limits_{x\to3^-}f(x)=\lim\limits_{x\to3^-}(x^2-1)=(3)^2-1=8$$

$$\lim\limits_{x\to3^+}f(x)=\lim\limits_{x\to3^+}(2bx)=2b(3)=6b$$

$$\lim\limits_{x\to3^-}f(x)=8=\lim\limits_{x\to3^+}f(x)=6b$$

$$b=\dfrac{4}{3}$$

Then

$$\lim\limits_{x\to3^-}f(x)=\lim\limits_{x\to3^+}f(x)=\lim\limits_{x\to3}f(x)=8$$

$$f(3)=2(\dfrac{4}{3})(3)=9$$

Then

$$\lim\limits_{x\to3}f(x)=8=f(3)$$

The function $f(x)$ is continuous everywhere if $b=\dfrac{4}{3} .$