is f(x)=(3x+4)^(1/2) continuos and differentiable

Question

anonymous · Answer

pretty much every function you know is continuous and differentiable on its domain

anonymous · Answer

$$f(x)=\sqrt{3x+4}$$ has domain 
$$x\geq -\frac{4}{3}$$ and it is continuous and differentiable  on the open interval $x>-\frac{3}{4}$
cannot include the left endpoint as it the limit doesn't exist from the left since the function is not defined for number less that $-\frac{3}{4}$

anonymous · Answer

so the mean value theorem would satisfyy it in the interval (0,4)?

anonymous · Answer

yes it would

anonymous · Answer

well actually "it would satisfy the mvt" not the other way around

anonymous · Answer

ok so my hw question is: "what value of c in the open interval (0,4) satisfies the mean value theorem for f(x)= (3x+4)^(1/2)

anonymous · Answer

The mean value theorem says there is a c in (0,4) such that f'(c) is the same as the average slope, which in this case is one half, by$$m_{average}=\frac{f(4)-f(0)}{4-0}=\frac{4-2}{4}=\frac{1}{2}$$Calculate f'(x), let f'(c)=1/2, and solve for c.