For all real $x$ the expression$ \frac{x^2-2x+4}{x^2+2x+4}$ lies between $\frac 13$ and $3$.

89 Views Asked by At

The values between which the expression $$\frac{9.3^{2x}+6.3^x+4}{9.3^{2x}-6.3^x+4}$$ is

A solution I saw says that the expression can be written as $$\frac{(3.3^x)^2+2(3.3^x)+4}{(3.3^x)^2-2.(3.3^x)+4}$$ which is analogous to the previous expression and hence would have the same interval. Why are they analogous? Clearly it has $x$ as a power, so something has to change right? Why are they the same?

1

There are 1 best solutions below

0
On BEST ANSWER

$$\frac{9.3^{2x}+6.3^x+4}{9.3^{2x}-6.3^x+4}$$

$$\frac{(3)^2(3^x)^2+2(3)(3^x)+4}{(3)^2(3^x)^2-2(3)(3^x)+4}$$

$$\frac{(3.3^x)^2+2(3.3^x)+4}{(3.3^x)^2-2(3.3^x)+4}$$

$$\frac{(z)^2+2(z)+4}{(z)^2-2(z)+4}$$