## Calculus: Early Transcendentals (2nd Edition)

$a.$ This is true. $b.$ This is true. $c.$ Not possible.
$a.$ The function is continuous if the small change of $x$ corresponds to the small change of the value of the function i.e. $$\lim_{h\to0}(f(x+h)-f(x))=0.$$ For the derivative to exist at $x$ we need that the following limit exists: $$f'(x)=\lim_{h\to0}\frac{f(x+h)-f(x)}{h}.$$ Since $h$ in the denominator tends to zero the only possibility for existance of this limit is that the numerator also tends to zero i.e. $$\lim_{h\to0}(f(x+h)-f(x))=0.$$ $b.$ This is true. This function is continuous because $$\lim_{h\to0}(f(x+h)-f(x))=\lim_{h\to0}(|x+h+1|-|x+1|)=|x+0+1|-|x+1|=0.$$ But it is not differentiable at $x=-1$ because there it haw two different tangents from two sides: For $x-1$ then $|x+1|=x+1$ so the slope of the tangent from the right is $1$. $c.$ The derivative of the function is defined on the interior of the interval and never on its endpoints. This means that the domain of the derivative is narrower than the domain of the function.