## University Calculus: Early Transcendentals (3rd Edition)

$g(x)$ is continuous at $0$ at $b=1$, but there is no value of $b$ that can make $g(x)$ differentiable at $0$.
$g(x)=x+b$ for $x\lt0$ and $g(x)=\cos x$ for $x\ge0$. 1) $g(x)$ is continuous at $0$ if and only if - $\lim_{x\to0}g(x)$ exists. - $g(0)$ exists (already satisfied). - $\lim_{x\to0}g(x)=g(0)=\cos0=1$ First, we need to find $\lim_{x\to0}g(x)$, by finding $\lim_{x\to0^+}g(x)$ and $\lim_{x\to0^-}g(x)$ for 2 intervals separately: $$\lim_{x\to0^+}g(x)=\lim_{x\to0^+}(\cos x)=\cos0=1$$ $$\lim_{x\to0^-}g(x)=\lim_{x\to0^-}(x+b)=0+b=b$$ For $\lim_{x\to0}g(x)$ to exist, it is necessary that $$\lim_{x\to0^+}g(x)=\lim_{x\to0^-}g(x)$$ $$b=1$$ And with $b=1$, $\lim_{x\to0}g(x)=1$. It is automatically equal with $g(0)$, making $g(x)$ continuous at $x=0$. 2) For $g(x)$ to be differentiable at $0$, first it must be continuous at $0$, which means $b=1$ is essential. So we need to test whether with $b=1$, $g(x)$ is differentiable or not. - For $x\lt0$: $g(x)=x+1$ $$g'(x)=(x+1)'=1$$ So, $$g'(0)=1$$ - For $x\ge0$: $g(x)=\cos x$ $$g'(x)=(\cos x)'=-\sin x$$ So, $$g'(0)=-\sin0=0$$ These two values are not equal, meaning that $g(x)$ is not differentiable for $b=1$. This only means there is no value of $b$ that can make $g(x)$ differentiable at $x=0$.