Find the slope of the linear function, $f$, where for all $x\in\mathbb{R}$, $ f(x-3)=f(x)+24$

Question

I’ve tried inverting both sides of the equation but I’m kind of stuck

Answer 1

The slope of a line connecting two points $(x_{1},y_{1})$ and $(x_{2},y_{2})$ is $\frac{y_{2}-y_{1}}{x_{2}-x_{1}}$. Now, plugging in $x=3$, you know that $f(0)=f(3)+24$. Hence, $f(3)-f(0)=-24$. Therefore, the slope is: $$ \frac{f(3)-f(0)}{3-0}=\frac{-24}{3}=-8. $$

Answer 2

If $f$ is a linear function, the $f(x) = mx + b$, where $m$ and $b$ are real numbers. The number $m$ represents the slope, which we get by changing $x$ slightly, then looking at how the value of the function (i.e. the $y$-value) changes in correspondence. They typical formula is $$ m = \text{slope} = \frac{y_2 - y_1}{x_2 - x_1}, $$ where $(x_1, y_1)$ and $(x_2, y_2)$ are two points that the graph of the function pass through. This is the change in $y$ over the change in $x$.

In this problem, every point on the graph is of the form $(x, f(x))$. Thus if we fix some $a$, we know that there is a point $(a,f(x))$ on the graph. Another point on the graph is given by $(a-3, f(a-3))$ (since we can choose any $x$-value, plug that into $f$, and get another point on the graph). Thus we can take $$ (x_1, y_1) = (a, f(a)) \qquad\text{and}\qquad (x_2, y_2) = (a-3, f(a-3)).$$ By hypothesis, $f(a-3) = f(a) + 24$, so we can write $y_2 = f(a) - 24$. Summarizing all of the above, we now have \begin{align} x_1 &= a & y_1 &= f(a) \\ x_2 &= a-3 & y_2 &= f(a) + 24. \end{align}

Plugging these into the above formula for slope, we finally obtain $$\text{slope} = m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{f(a-3) - f(a)}{(a-3) - (a)} = \frac{(f(a) + 24) - f(a)}{-3} = -\frac{24}{3} = -8. $$

Observe that ervx's (very fine) answer uses exactly the same approach, with the choice of $a=0$.