Explicitly write out the Lagrange interpolating polynomial of order $3$ for the data: $f(1)=0\mathrm{.}1924,$

$f(1\mathrm{.}05)=0\mathrm{.}2414,f(1\mathrm{.}10)=0\mathrm{.}2933,f(1\mathrm{.}15)=0\mathrm{.}3492,$ Do not expand.

Rather helpful note on the phrase Do not expand. I will not grade that aspect$($s$)$ of a problem which is

tantamount to asking for lost minus signs. Might even zap you a point or two. I know you are better than a

TI$-92,$ Voyage $200,$ TI$-$NSPIRE CAS, Maple or Mathematica $($some of the better known CAS tools$)$ thus

evidence of that likely will not impress me very much. Nor do I have the energy to plow through more

than my own efforts at such fun. More to the point, for something like a Lagrange interpolating

polynomial that takes a page to write down, one evaluates them numerically, Matlab has the capability to

do vector arithmetic thus the computation can be done for a lot of $x$ numbers at one time, less painful. The

book shows you haw to shortcut that process for computational purposes. So please to not worry about

wading through the algebra on such things. It is more sensible in a numerical methods class to evaluate

and plot your construction, i$.$e$.,$ implement visualization, to assess if a creation is likely to be correct.

Note on problems $1-4$: If requested to plot $f(x)$ and the interpolating polynomial together, also plot the

actual data pairs on the plots using a symbol such as o or $*$ to denote the points $($see help plot for how to

do this$),$ not broken line plots, but the points separately. The data is where an interpolating polynomial

will be exact. e$.$g$.,$ plot$($x$,$y$,$datax,datay, $\text{'}*\text{'})$ where $x,y$ are all the multitude of $x$ values at which to

evaluate your interpolation polynomial and the y values come from evaluation of the interpolating

polynomial at $x$; datax and datay are the data ordered pairs.