Where am i going wrong in solving this equation?

Question

Fing the least value of $a$ for which $f(x)$ is increasing, where $$f(x)=2e^x-ae^{-x}+(2a+1)x-3$$

What i tried

for increasing $f'(x)\ge 0, \forall x\in \mathbb R$. So $$f'(x)=2e^x+ae^{-x}+2a+1\geq0$$ and multiplying through by $e^x$ we get$$f'(x)=2e^{2x}+a+(2a+1)e^x\ge0$$and now put $y=e^x$

$$\implies2y^2+(2a+1)y+a\ge0$$ So discriminant ($D$) of this equation should be less than $0$ $$D=4a^2-4a+1\leq 0\implies(2a-1)^2\leq0$$which is only true when $a=1/2$, but this is not the correct answer! The least value of $a$ required is $0$ and $f(x)$ is increasing for $a\in[0,\infty[$ but my equation says it increases only if $a=1/2$

So where am i going wrong? Thanks in advance!

Answer 1

So where am i going wrong?

In the following part :

$$\implies2y^2+(2a+1)y+a\ge0$$ So discriminant ($D$) of this equation should be less than $0$

Note that $y\ (=e^x)$ is positive. So, what we want is the condition for $a$ that $2y^2+(2a+1)y+a\ge 0$ holds for every positive $y$. (That you had $D\le 0$ means that you found the condition for $a$ that $2y^2+(2a+1)y+a\ge 0$ holds for every $y$, including non-positive $y$).

We can write $$F(y):=2y^2+(2a+1)y+a=2\left(y+\frac{2a+1}{4}\right)^2-\frac{(2a-1)^2}{8}$$

The vertex of the parabola $Y=F(y)$ is $(-(2a+1)/4,-(2a-1)^2/8)$ where $-(2a-1)^2/8\le 0$.

We have to have $$-\frac{2a+1}{4}\lt 0\quad \text{and}\quad F(0)\ge 0\iff a\ge 0$$ and this is sufficient.

$\qquad\qquad\qquad$

Answer 2

So you want $f'(x) \ge 0$ for all $x$, which led to: $$2e^x+ae^{-x}+2a+1 \ge 0 \iff 2e^{2x}+\left(2a+1\right)e^x+a\ge 0$$ If you factor this: $$\left(2e^x+1\right)\left(e^x+a\right)\ge 0$$ The first factor is always positive and $e^x+a \ge 0$ for all $x$ if $a \ge 0$.

Answer 3

Notes:
1) $y>0$ everywhere;
2) when $D<0,$ then $y>0$ everywhere.

Factorization is the right way, and we have to resolve inequality $$f'(x) = e^{-x}(2e^x+1)(e^x+a)\geq0,$$ $$e^x+a\geq 0.$$ If $a\geq0,$ then $f(x)$ increases everywhere.
If $a<0$, then $f(x)$ increases with $x\in(\ln(-a),\infty).$