finding the minimum slope of a tangent line Given a tangent line that touches the function $f(x)=\frac{1}{\sqrt{18-2x}}$ at $x=a$, $(0\le a\le 7)$.
Need to find the value of $x=a$ when the slope of the tangent line will be minimum.
I found that there are no $x$ such that $f(x)'=0$ so i need to check the endpoints, but where do i check them? in $f(x)$ or $f'(x)$?
Thanks.
 A: Calculating the formula for the $f'(x)$, we have $f'(x)=(18-2x)^{-3/2}$.   Note that in our domain of $[0,7]$,  as x increases,  $18-2x$  decreases,  then raising that to a negative power makes it increase,  hence $f'(x)$  is an increasing function on $[0,7]$  Thus, it will be minimized at the beginning of the domain,  at $x=0$.
A: So you know that $f'(x)$ is the slope. Now what you want to do is minimise the slope. This is where you should think: minimise means taking the derivative and setting it equal to zero, then figuring out which zeros of the derivative are minima. So you want to find $f''(X)$ and set it equal to zero, and then determining its minimum from that.
A: Since $f(x)=\frac{1}{\sqrt{18-2x}}$ then $f'(x)=\frac{1}{(18-2x)^{3/2}}$.  Now, we need to examine the second derivative to see if this function is monotonic on $[0,7].$  We have
$$f''(x)=\frac{3}{(18-2x)^{5/2}}$$ which means $f(x)$ is monotonic for all $x$ in the domain since $f''(x)$ is never equal to $0$.  Therefore, it is sufficient to check only the endpoints and plug them into $f'(x)$.  Plugging $x=0$ into $f'(x)$ we have
$$f'(0)=\frac{1}{18^{3/2}}\approx .0131$$
and plugging $x=7$ into $f'(x)$ we have
$$f'(7)=\frac{1}{4^{3/2}}\approx .125.$$
Since $f'(0)<f'(7)$ then we can conclude $f(x)$ has minimum slope of tangent line when $x=0.$
