Find slope at (4,2) and equation of tangent line

Question

given $$f(x)=\sqrt{x}, p = (4,2)$$

The formula for the slope is $$\frac{f(x-h)-f(x)}{h}$$

Which gives me $$\frac{\sqrt{4-h}-\sqrt{4}}{h}$$ which gives me $$\frac{\sqrt{4-h}-2}{h}$$

I know m = 1/4 but I'm not sure how to get there from my last step. Once I know the slope the rest is not difficult.

Answer 1

From the slope formula, you should actually have $\displaystyle \frac{\sqrt{4+h}-2}{h}$.

Now, let's make $h$ super small, and this value should approach the slope of the tangent line.

We have $\displaystyle \frac{\sqrt{4+h}-2}{h} = \frac{\sqrt{4+h}-2}{h} \cdot \frac{\sqrt{4+h}+2}{\sqrt{4+h}+2}=\frac{(4+h)-2^2}{h(\sqrt{4+h}+2)}=\frac{h}{h(\sqrt{4+h}+2)}$.

Now, this is equal to $\displaystyle \frac{1}{\sqrt{4+h}+2} $. We see that as $h$ grows small, the slope should approach $\displaystyle \frac{1}{\sqrt{4}+2}=\frac{1}{4}.$

Answer 2

What you've calculated is the opposite of the slope of the chord defined by the points $(4,2)$ and $(4-h,\sqrt{4-h})$.

The slope of the tangent line at a point on a (differentiable) curve is simply the value of the derivative at this point, so $$m=\frac 1{2\sqrt x}\Biggr|_{x=4}=\frac14.$$

As to the equation of the tangent line, you have a formula for that: $$y-f(x_0)=f'(x_0)(x-x_0).$$

Added (as the O.P. hasn't seen derivatives yet):

The tangent line at $(4,2)$ is the limiting position of a chord through this point and another point $(4-h,\sqrt{4-h})$ when the latter tends to $(4,2)$, i.e. when $h$ tends to $0$. So to obtain the slope of the tangent, we need to determine the limit of the slopes of the chords: $$\frac{\sqrt{4+h}-2}h, \quad\text{or, if you prefer},\quad \frac{\sqrt{4-h}-2}{-h}.$$ To obtain the limit, we use the trick of the conjugate expression to rationalise the numerator: $$\frac{\sqrt{4-h}-2}{-h}=\frac{(\sqrt{4-h}-2)(\sqrt{4-h}+2)}{-h(\sqrt{4-h}+2)}=\frac{\not4-h-\not4}{-h(\sqrt{4-h}+2)}=\frac{1}{(\sqrt{4-h}+2)},$$ which tends to $\;\frac1{\sqrt{4\,}+2}=\frac14$.