The function $f$ is linear and satisfies $f(d+1)-f(d) = 3$ for all real numbers $d$. What is $f(3)-f(5)$?
Taking $d = 3,$ we get
\[f(4) - f(3) = 3.\]Taking $d = 4,$ we get
\[f(5) - f(4) = 3.\]Adding these equations, we get $f(5) - f(3) = 6,$ so $f(3) - f(5) = \boxed{-6}.$
-6
Level 2
Intermediate Algebra
Problem: The function $f$ is linear and satisfies $f(d+1)-f(d) = 3$ for all real numbers $d$. What is $f(3)-f(5)$?
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To find the value of $f(3) - f(5)$, we first need to understand the implications of the condition $f(d+1) - f(d) = 3$. This indicates that the function $f$ has a constant difference of 3 between consecutive integer inputs, suggesting that $f$ is a linear function of the form $f(x) = mx + b$ where $m$ is the slope and $b$ is the y-intercept.

Given that $f(d+1) - f(d) = 3$, we can express this in terms of the linear function: 

f(d+1) = m(d+1) + b = md + m + b
f(d) = md + b

Subtracting these two equations gives:

f(d+1) - f(d) = (md + m + b) - (md + b) = m = 3.

Thus, the slope $m$ of the function $f$ is 3. Therefore, we can express the function as:

f(x) = 3x + b, where $b$ is some constant.

Next, we need to calculate $f(3)$ and $f(5)$:

f(3) = 3(3) + b = 9 + b,
f(5) = 3(5) + b = 15 + b.

Now, we can find $f(3) - f(5)$:

f(3) - f(5) = (9 + b) - (15 + b) = 9 + b - 15 - b = 9 - 15 = -6.

Thus, the final answer is $-6$.<|END_OF_LATENT|> The final answer is $-6$.