How to prove that function-value is zero before specific input?

Question

I have been asked to prove that

$$ f(x)=2x-ae^{-x}(x^{2}+1) \; \; \text{where} \;\; a>0 $$

will reach the value

$$ f(x_{0})=0 \;\; \text{when} \;\;x_{0}<\frac{a}{2} $$

but have noe clear idea about how to proceed. What should I do?

Answer 1

From the Taylor series for the exponential, we know that $$e^x\ge 1+x+\frac12x^2\qquad \text{for }x\ge0. $$ Therefore, for $x\ge0$, $$e^{-x}(1+x^2)=\frac{1+x^2}{e^x}\le \frac{1+x^2}{1+x+\frac12x^2}<\frac{2+2x+x^2}{1+x+\frac12x^2}=2. $$

We conclude using $a>0$ that $$f(\tfrac a2)=a-\frac a2e^{-\frac a2}\left(1+\frac{a^2}4\right)> a-\frac a2\cdot 2=0.$$ As $f(0)=-a<0$, the Intermediate Value Theorem tells us that $f(x_0)=0$ for some $x_0$ between $0 $ and $\frac a2$.