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.
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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 \(\frac{5}{2}\): \[ x = -1 \cdot \frac{5}{2} = -\frac{5}{2} \]
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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:
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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: \[ y = f(-1 + 4) + f(-\frac{15}{2} - 1) = f(3) + f(-\frac{17}{2}) = 4 + f(-\frac{17}{2}) \]
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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.