Science:Math Exam Resources/Courses/MATH100/December 2010/Question 07/Solution 1

Taylor's remainder formula says that

$\displaystyle |T_{2}(1)-f(1)|={\frac {|f'''(c)||x-a|^{3}}{3!}}\leq {\frac {M|1-0|^{3}}{6}}={\frac {M}{6}}$

where M is an upper bound for the absolute value of the third derivative on the interval $\displaystyle [0,1]$. Computing the third derivative via the product rule gives

${\begin{aligned}f(x)&=e^{x}(x^{2}-7x+15)\\f'(x)&=e^{x}(x^{2}-7x+15)+e^{x}(2x-7)\\&=e^{x}(x^{2}-7x+15+2x-7)\\&=e^{x}(x^{2}-5x+8)\\f''(x)&=e^{x}(x^{2}-5x+8)+e^{x}(2x-5)\\&=e^{x}(x^{2}-5x+8+2x-5)\\&=e^{x}(x^{2}-3x+3)\\f'''(x)&=e^{x}(x^{2}-3x+3)+e^{x}(2x-3)\\&=e^{x}(x^{2}-3x+3+2x-3)\\&=e^{x}(x^{2}-x)\\\end{aligned}}$

Notice that $\displaystyle e^{x}$ is increasing on $\displaystyle [0,1]$ and thus obtains its maximum at x = 1, i.e.

$|e^{x}|=e^{x}\leq e^{1}=e$

for $x\in [0,1]$.

The function $\displaystyle x^{2}-x$ is a parabola with roots at 0 and 1. Hence its minimum occurs at the midpoint of the roots, namely at $\displaystyle x=1/2$. Since the value at the endpoints x = 0 and x = 1 is zero, the minimum value of the parabola maximizes its absolute value. Thus,

$\displaystyle |x^{2}-x|\leq |(1/2)^{2}-(1/2)|=1/4$

Since we want to bound the absolute value of the third derivative, we know that

$\displaystyle |f'''(x)|=|e^{x}||x^{2}-x|\leq e\cdot 1/4=e/4=M$.

Hence, we have

$\displaystyle |T_{2}(1)-f(1)|\leq {\frac {M}{6}}={\frac {e}{24}}$

as required.