So we have the slope of our tangent line to be the slope of our function.
$$f'(x) = \frac{1}{x^2}$$
At some point on $f(x)$ and our tangent, which we can denote as $(c, \frac{c-1}{c})$
Our slope now becomes:
$$f'(c) = \frac{1}{c^2}$$
Now we have a point for our tangent line, $(4, 1)$ and another point $(c, \frac{c-1}{c})$
So let us use our point-slope form:
$$y - 1 = m(x-1)$$
$$y - \frac{c-1}{c} = \frac{1}{c^2}(x-c)$$
Now we can use our point $(4, 1)$
$$1 - \frac{c-1}{c} = \frac{1}{c^2}(4-c)$$
Solving we find that $c = 2$
So, our tangent line is
$$y - \frac{2-1}{2} = \frac{1}{4}(x-2)$$
$$y =\frac{x}{4}$$
We find the equation of the tangent line to $f(x) = x^3-x$ at the point $(k, k^3-k)$.
First, the derivative gives the slope
$$f'(x) = 3x^2 -1$$
So we have a line with slope $3k^2-1$ that goes through the point $(k,k^3-k)$. What is its equation? We use the point-slope form
$$y-(k^3-k) = (3k^2-1)(x-k)$$
$$y-k^3 +k = 3k^2x-x-3k^3+k$$
$$y+2k^3 = 3k^2x-x$$
We want to know which values of $k$ make this line pass through $(x,y)=(-2,2)$, so we plug in those values:
$$2+2k^3=3k^2(-2)-(-2)$$
$$2k^3 = -6k^2$$
$$k^3 = -3k^2$$
$$k=0 \,\,\,\,\text{ or }\,\,\, k=-3$$
This means that there are two tangent lines that work, one at $(0,0)$, and the other at $(-3,-24)$. You should be able to find the equations of these lines easily if necessary.
Best Answer
At the point $\left(t,(t+1)^{\frac 32}\right), \ f'(t)= \frac 32 \sqrt{t+1}$. The equation of the tangent to the curve at that point is $y = \frac 32 \sqrt{t+1}(x - t) + (t+1)^{\frac 32}$.
Now let $x=\frac 43$and $y=3$. Then solve for $t$. You should get two nice integers for answers.
\begin{align} \frac 32 \sqrt{t+1}\left(\frac 43 - t \right) + (t+1)^{\frac 32} &= 3\\ \sqrt{t+1}(4-3t) + 2(t+1)^{\frac 32} &= 6 \\ (4-3t) + 2(t+1) &= \frac{6}{\sqrt{t+1}} \\ 6-t &= \frac{6}{\sqrt{t+1}} \\ etc \end{align}