Find values of parameter $m$ for which $m9^x + 4(m-1)3^x+m>1$ 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.
 A: 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.
A: Note: I wrote this answer before my comments above were answered. This answer is based on the assumption that the ranges given for the solution options are for the variable $m$.
for:
$$ f(m,x)= m9^x + 4(m-1)3^x+m \tag0$$ 
as I understand, we need to find the domain for m which makes $f(m,x)>1$.
Regarding your solution, I assume that it follows a correct path, but the solution of:
$$mt^2\:+4\left(m-1\right)t+m-1=0 
\tag1$$
is (for m not equal zero):
$$t=\:\frac{-2m+2\sqrt{\left(3m^2-7m+4\right)}}{m},\:\:\frac{-2m-2\sqrt{\left(3m^2-7m+4\right)}}{m}$$
Which is different from what you have.
The expression in (1) has real values for $t$ when $m=\frac{4}{3}$ or $m=1$.
You now need to check $f(1,x)$ ,$f(\frac{4}{3},x)$. This is rather lengthy and may not be easy.
Another approach, probably simpler, is to remove invalid options as follows:
When m=$0$, 
$$f(0,x)= -4(3^x)$$ 
The above value is always negative, so answer (E) is excluded. If (E) is wrong, then (A) is wrong.
Now you have to choose between (B), (C)  and (D).
Given the way option (C) is written, it encloses infinity, so it is an impossible option. As a result we need to consider (B) or (D).
Considering (B), Test $m=1$, to get:
$$f(1,x)= (9^x)+1$$ 
When $f(1,x)>1$:
$$(9^x)+1>1$$ 
That is:
$$(9^x)>0$$ 
The above is always true. 
Since you have to pick only 1 answer, then (B) is correct and there is no need to test (D) since (D) must be wrong if only one answer is correct.
Edits Made: Due to discussion below I made 2 edits regarding the invalidity of (C) and (D).
A: Long answer first, by direct calculation, then a short answer, by deduction.
You have two mistake in your solution.  Fisrt, the expression
$$f_m(t)=mt^2+4(m-1)t + m-1$$
need to be strickly positive for every $x\in\mathbb{R}$. Since $t = 3^x$, it need to be true only for $t \in \mathbb{R}^+$.
$m>0$, so the parabola is positive after the rightmost root. And this rightmost root can't be greater than $0$. The parabola can be $\leq0$ as long as it is for $t\leq 0$.
Second, you made a mistake in the calculation of $\Delta$.
$$\Delta = 4(m-1)(3m-4)$$
$\Delta<0$ for $m\in\left(1,\dfrac43\right)$, those values of $m$ are accepted.
If $m=1$, $f_1(t) = t^2$, it is a parabola with vertex at $(0,0)$. $f_1(t)>0$ if $t>0$, it is an accepted value.
For other values of $m$, the roots of $f_m(t)$ are given by
$$\dfrac{-4(m-1)\pm \sqrt{4(m-1)(3m-4)}}{2m} = \dfrac{4-4m\pm\sqrt{4(m-1)(3m-4)}}{2m}$$
Sine $m>0$, the rightmost root is
$$\dfrac{4-4m+\sqrt{4(m-1)(3m-4)}}{2m}$$
If $m<1$, this expression is positive, so every value of $m<1$ is not part of the solution.
If $m\geq \dfrac43$, we need to show that 
$$\dfrac{4-4m+\sqrt{4(m-1)(3m-4)}}{2m}\leq 0$$
Since $3m-4 < 4m-4$, 
$$\dfrac{4-4m+\sqrt{4(m-1)(3m-4)}}{2m}\leq \dfrac{4-4m+\sqrt{4(m-1)(4m-4)}}{2m}= 0$$
Every value of $m\geq \dfrac43$ is an accepted value
Answer B. $[1,+\infty)$.

Short answer
Since it is a multiple choices question, we can find the answer by deduction. A lot of answer have $m = 1$ as boundary, so we'll check if it work.
$$(1)\cdot 9^x+4((1)-1)3^x+ (1) = 9^x+1 >0 \qquad x\in\mathbb{R}$$
$m=1$ is part of the solution. Possible answer B. or C.
Does it work for $m=\dfrac12$?
$$\left(\dfrac12\right)\cdot9^x+4\left(\left(\dfrac12\right)-1\right)3^x+\left(\dfrac12\right)=\dfrac{9^x-4\cdot3^x+1}{2}$$
If we take $x = -1$ then
$$\dfrac{9^{-1}-4\cdot3^{-1}+1}{2}=-\dfrac19 < 1$$
$m=\dfrac12$ is not part of the solution, which rule out answer C.
Thus B. is the answer.
