The exercise asks for the $\bar x$ value as $f(\bar x) = 4.5$ using linear interpolation.
I couldn't reproduce a table with MathJax, so I'll put the tabled with data as:
$f(2) = 10$
$f(4) = 13$
$f(5) = 6$
$f(7) = 5$
$f(9) = 2$
Since I have $f(\bar x) = 4.5$ then my $\bar x$ value must be around $f(7) = 5$ and $f(9) = 2$.
Using linear interpolation, I found $P_{1}(x) = -1.5x - 5.5$ Now, replacing $P_{1}(x)$ for 4.5, I'll get $4.5 = -1.5x -5.5$ then $x = 6.666667$
But the answer is $7.333333$
I was expecting having to sum the $x$ value with an error number, but I couldn't find a way to calculate the error as well. Am I missing something?
$\bar{x}= \frac{(4.5-5)}{(2-5)}(9-7) + 7 = \frac{-0.5}{-3}\cdot2+7=7.333333$