I am asked to find all values of $m$ for which:
$$m9^x + 4(m-1)3^x+m>1, \forall x \in \mathbb{R}$$
The possible solutions are:
A. $(- \infty, 1)$
B. $[1, + \infty)$
C. $(0, + \infty)$
D. $(1, + \infty)$
E. $\emptyset$
I got to a point and then I got stuck. This is what I tried:
$$m9^x + 4(m-1)3^x+m>1$$
$$m9^x + 4(m-1)3^x+m – 1 > 0$$
Here I used the substitution:
$$t = 3^x, t > 0$$
And then the expression became:
$$mt^2 + 4(m – 1)t + m – 1 > 0$$
I know that this is true iff:
$(I). m > 0$. This is because we need a convex parabola for the expression to be $>0$ $\forall x \in \mathbb{R}$.
$(II). \Delta = b^2-4ac$ must be $<0$. This way the parabola is above the $Ox$ axis in its entirety, therefore always positive.
$$\Delta = 16(m-1)^2-4m(m-1)$$
$$\Delta=4(m-1)[4(m-1)-m]$$
$$\Delta=4(m-1)(3m-1)$$
Solving $\Delta < 0$, I got:
$$m \in \bigg (\dfrac{1}{3}, 1 \bigg )$$
Combining this with the condition from $(I)$ I get:
$$m \in \bigg ( \dfrac{1}{3}, 1 \bigg)$$.
So my conclusion is that for $m \in \bigg (\dfrac{1}{3}, 1 \bigg )$, I have:
$$mt^2 + 4(m – 1)t + m – 1 > 0, \forall t \in \mathbb{R}$$
But these are not the solutions for the initial expression. $m \in \bigg ( \dfrac{1}{3}, 1 \bigg )$ are the solutions for the transformed expression, where $t = 3^x$. So how can I translate these values of $m$ for which the transformed, $t$-based expression is true into an interval of values of $m$ for which the initial, $x$-based expression is true? Obviously, this interval would have to be among the $5$ given possible answers A, B, C, D or E.
Or, alternatively, if my approach seems inefficient or flat out wrong, I'd appreciate it if you could guide me towards the right approach.
Best Answer
Here is a straightforward approach. Rearrange the inequity as
$$m > \frac{4\cdot 3^x+1}{9^x+4\cdot 3^x+1}=\frac{1}{1+\frac{3^{2x}}{4\cdot 3^x+1}}$$
The RHS is a sigmoid function with the range $(0,1)$. Thus, the answer (B), $[1,\infty)$, always satisfies the inequity.