Problems need HardWork 20

First we can see that for all $x \in R^{+}$ :

$6^{x}>3^{x}>x^{2}>0$ and $2^x>0$

The unobviously correct part is $3^{x}>x^2$ . Note that we still have $x \in R^{+}$ . For proving this first we need to take the first and second derivative of $3^x-x^2$ :

$\frac{d}{dx}(3^x-x^2)=3^x \ln 3 - 2x$

$\frac{d}{dx}(3^x \ln 3 - 2x) = 3^x \ln^2 3 - 2$

Set y as the value of x that make the value of the second derivative equals 0:

$y = \log_3 \frac{2}{\ln^2 3}$

Notice that whenever $x>y$ , the second derivative becomes positive, and whenever $x<y$ , the second derivative becomes negative. This means the first derivative is minimum at $x=y$ , and that minimum equals:

$\frac{2}{\ln 3}-2 \log_3 \frac{2}{\ln^2 3} = \frac{2}{\ln 3}(1-\ln \frac{2}{\ln^2 3}) > 1-\ln 2 > 0$

because $0<\ln 2<1<\ln 3<2$ . Thus, the first derivative is always positive.

Since $3^0>0^1$ and the first derivative is always positive, then it's proven that $3^x>x^2$ for all $x \in R^{+}$ .

Hence, all solutions of x lies in $R^{-}$ as x=0 is not a solution, obviously. Now, $f(x)=2^x+3^x+6^x$ is a monotone increasing function at all values of $x \in R$ while $g(x)=x^2$ is a monotone decreasing function at all values of $x \in R^{-}$ . As a result, there is a maximum of one value of x in $R^{-}$ that satisfy the equation. As there is no solution of x in $[0,\infty)$ and $x=-1$ does satisfy the equation, then the only solution is $\boxed{x=-1}$ . Any feedback please?

First prove that

f (x)= 2^(x) + 3^(x) + 6^(x) - x^(2)

is a strictly increasing function. This can be done by using the first derivative and the second derivatives. Then is proves that -1 is the only root.

$2^{x+1}+3^{x+1}+6^{x+1}$ increases the value of $2^{x}+3^{x}+6^{x}$ by $2^{x}+3^{x}+3^{x}$ +... which is greater than 2x+1 So LHS would be always greater than RHS where x>0 So the cases left are 0 and negative nos. x can't be 0 after verification. Now the negative values : -1 satisfies the condition.Then the denominators later on(onwards 2) can't be completely divisible by the numerator.So only -1 can be the answer.