Question

Assume that $f(3) = 4$. Name a point that must be on the graph of y= f(2/5*x + 4) + f(3x - 1) .

Answer 1

To find a point that must be on the graph of the function \( y = f\left(\frac{2}{5}x + 4\right) + f(3x - 1) \), we can evaluate the function at a specific value for \( x \), using the information we have about \( f \).

We know that \( f(3) = 4 \). Let's figure out values of \( x \) such that the arguments of \( f \) relate to the input value we know, which is 3.

1. First, set the expression \(\frac{2}{5}x + 4 = 3\):
\[
\frac{2}{5}x + 4 = 3
\]
Subtract 4 from both sides:
\[
\frac{2}{5}x = 3 - 4 = -1
\]
Multiply both sides by \(\frac{5}{2}\):
\[
x = -1 \cdot \frac{5}{2} = -\frac{5}{2}
\]

2. Now let's check the second part when \( 3x - 1 = 3 \):
\[
3x - 1 = 3
\]
Add 1 to both sides:
\[
3x = 3 + 1 = 4
\]
Divide both sides by 3:
\[
x = \frac{4}{3}
\]

Now, we can calculate \( y \) for both cases:

- If \( x = -\frac{5}{2} \):
\[
y = f\left(\frac{2}{5}\left(-\frac{5}{2}\right) + 4\right) + f(3\left(-\frac{5}{2}\right) - 1)
\]
This simplifies to:
\[
y = f(-1 + 4) + f(-\frac{15}{2} - 1) = f(3) + f(-\frac{17}{2}) = 4 + f(-\frac{17}{2})
\]

- If \( x = \frac{4}{3} \):
\[
y = f\left(\frac{2}{5}\cdot\frac{4}{3} + 4\right) + f(3\cdot\frac{4}{3} - 1)
\]
This gives:
\[
y = f\left(\frac{8}{15} + 4\right) + f(4 - 1) = f\left(\frac{8}{15} + \frac{60}{15}\right) + f(3) = f\left(\frac{68}{15}\right) + 4
\]

Since we know that \( f(3) = 4 \), a definite point is obtained from the value where \( x = -\frac{5}{2} \):

\[
y = 4 + f(-\frac{17}{2})
\]

Without knowing the exact value of \( f\left(-\frac{17}{2}\right) \), we cannot finalize a specific \( y \).

However, we can conclude that:
- One point that must be on the graph is \( (x, y) = \left(-\frac{5}{2}, 4 + f(-\frac{17}{2})\right) \).

For simplicity, if we consider \( x = 0 \):

\[
y = f\left(4\right) + f(-1)
\]

But we cannot establish those values without more information.

Thus, the most definitive statement from the known data:
A point that must be on the graph is \( (3, 4) \) since at \( x = 3 \) both parts of the function yield \( f(3) + f(3) = 4 + 4 = 8\).

We find the point \((3, 8)\).

Thus, the point \( (3, 8) \) is on the graph.