Translating a function vertically I apologize if this is obvious. I'm trying to gain an intuition on plotting functions. The example I have to work with is comparing the graph of $f(x) = \sqrt{x}$ and $g(x) = 1 + \sqrt{x-1}$. I know what $\sqrt{x}$ looks like. The graph of $\sqrt{x-1}$ is a horizontal translation by one unit, and the graph of $g(x)$ then shifts this function one unit upward.
My question is this: from plotting the two functions, I see that everywhere $\sqrt{x-1} < \sqrt{x}$, i.e., I could obtain $\sqrt{x}$ by "shifting" the function of $\sqrt{x-1}$ upward by a distance that is first somewhat wide and then appears to get constant. After working out the derivatives, I see that for any given $x$, the derivative of $\sqrt{x-1}$ is strictly greater than the derivative of $\sqrt{x}$, so I'd expect the graph of $\sqrt{x-1}$ to be somewhat steeper, but this isn't obvious to me from the plot.
I'd appreciate if someone could help with the intuition of these graphs.
 A: Ok, let's look at the point $(x,f(x))$ with $(x,f(x)) = (x, \sqrt{x}) $ .
Now let us look at the point that is one unit above $(x,f(x))$ . We can simply write it out as $(x,f(x)+1)$. This means for $h(x) := f(x)+1 = \sqrt{x} + 1$ the graph at any point $x$ is precisely one unit above $f(x)$ .
Now let's look at the point $(x+1, h(x))$ one to the right of $(x,h(x))$ . Again we can compute with $\tilde{x} := x+1$ :
$$
(x+1,h(x)) = (\tilde{x}, h(\tilde{x}-1)) = (\tilde{x}, \sqrt{\tilde{x} -1} +1) = (\tilde{x},g(\tilde{x}))
$$
Since this is true for any $\tilde{x}$ , we know that $ (x,g(\tilde{x}))$ is exactly shifted as described.
A: Let $\alpha(x)=\sqrt{x}$ and $\beta(x)=\sqrt{x-1}$. For any $x\ge1$, the tangent line to $\bigl(x,\beta(x)\bigr)$ is indeed steeper than the tangent line to $\bigl(x,\alpha(x)\bigr)$. However, the difference in steepness approaches $0$ as $x\to\infty$. This is because
\begin{align}
\lim_{x\to\infty}\beta'(x)-\alpha'(x) &= \lim_{x \to \infty}\frac{1}{2\sqrt{x-1}}-\frac{1}{2\sqrt{x}} \\[5pt]
&= \lim_{x \to \infty}\frac{1}{2\sqrt{x}} \left(\frac{\sqrt{x}}{\sqrt{x-1}}-1\right) \\[5pt]
&= 0 \cdot 0 \\[5pt]
&= 0 \, .
\end{align}
Hence, it shouldn't be obvious to you from the plot that the tangent lines are steeper—for large enough $x$, the difference is imperceptible!
