Science:Math Exam Resources/Courses/MATH100/December 2010/Question 07/Solution 2

Proceed as in solution 1 to see that

${\begin{aligned}f'''(x)&=e^{x}(x^{2}-x)\end{aligned}}$

We now need a bound on this function on $\displaystyle [0,1]$. Notice that the third derivative is neither monotonically increasing nor is it monotonically decreasing on the interval. Hence we are not guaranteed to find the maximum value by just looking at the endpoints. In fact, at the endpoints we have $\displaystyle f'''(0)=0=f'''(1)$, which is certainly not the maximum of the absolute value $\displaystyle |f'''(x)|$. Instead we do it the proper way: Take the derivative

$\displaystyle f''''(x)=e^{x}(x^{2}-x)+e^{x}(2x-1)=e^{x}(x^{2}+x-1)$

and set it to zero

$\displaystyle 0=e^{x}(x^{2}+x-1)$

Recall that $\displaystyle e^{x}\neq 0$. Using the quadratic formula, we see that

$\displaystyle x={\frac {-1\pm {\sqrt {1^{2}-4(1)(-1)}}}{2(1)}}={\frac {-1\pm {\sqrt {5}}}{2}}$

only the positive such root is in $\displaystyle [0,1]$. Let

$\displaystyle \phi :={\frac {-1+{\sqrt {5}}}{2}}$.

Then, our function $\displaystyle f'''(x)$ obtains an extreme value at $\displaystyle \phi$. Since the function is nonzero at this point, we know that $\displaystyle |f'''(x)|$ is maximal at $\displaystyle \phi$. Thus, we have that

$\displaystyle |T_{2}(1)-f(1)|\leq {\frac {M}{6}}={\frac {|f'''(\phi )|}{6}}={\frac {e^{\phi }|\phi ^{2}-\phi |}{6}}$

completing the solution.