Equation of the tangent to a graph where the point is not on the graph.

Question

If $f(x)=(x+1)^{3/2}$, provided $x\geq -1$, I am asked to find the equation of all tangent lines to $f(x)$ at the point $(\frac{4}{3},3)$.

Simple enough. I first took the derivative which is:

$$f'(x)= \frac {3\sqrt{x+1}}{2}$$

Since $(\frac{4}{3},3)$ is not on the graph of $f(x)$, I need to find the point of tangency.

So I assumed the point of tangency is at some point $(k,(k+1)^{\frac{3}{2}})$ It also has to pass the point $(\frac{4}{3},3)$ so thus:

$y-y_{1} = m(x-x_{1})$

$3-(k+1)^{\frac{3}{2}} = \frac{3}{2}\sqrt{k+1}(\frac{4}{3}-k)$

$3-\sqrt{(k+1)^2(k+1)} = 2\sqrt{k+1}-\frac{3}{2}k\sqrt{k+1}$

$3-(k+1)\sqrt{k+1} = 2\sqrt{k+1}-\frac{3}{2}k\sqrt{k+1}$

$3-k\sqrt{k+1}-\sqrt{k+1}=2\sqrt{k+1}-\frac{3}{2}k\sqrt{k+1}$

$3+\frac{1}{2}k\sqrt{k+1}-3\sqrt{k+1}=0$

Here is where I am stuck. How do I solve for $k$ (And hence $x$)?

I tried letting $\sqrt{k+1}$ equal $a$ and then try and solve the corresponding cubic but that's not getting me anywhere... Any help on this? Thanks!

Answer 1

$$3-\frac{1}{2} k\sqrt{k+1} - 3 \sqrt{k+1} =0$$ $$3=\frac{1}{2}k\sqrt{k+1} + 3 \sqrt{k+1}$$ $$3=\sqrt{k+1}(\frac{1}{2}k+3)$$ $$9=(k+1)(\frac{1}{4}k^{2}+9+3k)$$ $$\frac{1}{4}k^3 + \frac{13}{4}k^2+12k=0$$ $$k(\frac{1}{4}k^2 + \frac{13}{4}k+12)=0$$ from here you now one answer is $k=0$ and others are those who make $$\frac{1}{4}k^2 + \frac{13}{4}k+12=0$$ and with solving it you got $$\frac{-13\pm \sqrt{157}}{2}$$ as other answers but none of these satisfy $$\sqrt{k+1}>0$$ so your only answer will be k=0.

Answer 2

HINT

Use the fact that your line passes through $(k,f(k))$ and $(4/3,3)$, so the slope must also be $$ m = \frac{\Delta y}{\Delta x} = \frac{3 - f(k)}{4/3 - k} $$ and an alternate equation would be $$ y - 3 = m(x - 4/3) $$ but the equations have to match exactly...