Determine the equation of a non-piecewise function that satisfies the following condi

MaddieS

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Determine the equation of a non-piecewise function that satisfies the following conditions.
-Increases over the domain (x<-4)
-decreases over the domain (x>-4)
D=(xER / x(cannot equal) -4)
R=(yER/ y>3)
 
Determine the equation of a non-piecewise function that satisfies the following conditions.
-Increases over the domain (x<-4)
-decreases over the domain (x>-4)
D=(xER / x(cannot equal) -4)
R=(yER/ y>3)
 
help!

Determine the equation of a non-piecewise function that satisfies the following conditions.
-Increases over the domain (x<-4)
-decreases over the domain (x>-4)
D=(xER / x(cannot equal) -4)
R=(yER/ y>3)
 
help please

Determine the equation of a non-piecewise function that satisfies the following conditions.
-Increases over the domain (x<-4)
-decreases over the domain (x>-4)
D=(xER / x(cannot equal) -4)
R=(yER/ y>3)
 
The point (3,27) is on the function f(x)=3x. Determine its image point on the transformed function g(x)=0.4(3)2x-6-11. (The 2x-6 is exponential, the computer would not allow me to show this.)
It is standard to use "^" to indicate an exponent. That is "3^(2x- 6)" (note the entire exponent in parentheses for clarity) to indicate \(\displaystyle 3^{2x- 6}\). And, I presume, f(x)= 3^x, NOT "3x".

But I am puzzled as to exactly what the question is. There is no "f(x)" in the "transformed function", g(x). I guess you are talking about rewriting g(x) as \(\displaystyle 0.4 3^{2x}3^{-6}- 11= 0.4(3^{-6})(3^x)^2- 11= 0.4(3^{-6})(f(x))^2- 11\). Then \(\displaystyle 27^2= 729\) and \(\displaystyle 3^{-6}= \frac{1}{3^6}= \frac{1}{729}\) so that \(\displaystyle (3^{-6})(f(3))^2= (3^{-6})(27^2)= 1\) and then \(\displaystyle 0.4(3^{-6})(f(x))^2- 11= 0.4- 11= -10.6\).

But that's the hard way to do it! The obvious thing to do is to directly calculate \(\displaystyle (0.4)3^{6- 6}- 11= 0.4- 11= -10.6 since any (non-zero) number to the 0 power is 1.\)
 
Notice that, while the problem says "find the equation", it says of a function. There are an infinite number of functions satisfying these conditions and you just have to find one of them. The fact that the domain is "all real numbers except -4" makes me think of a fraction having x+ 4 in the denominator. The fact that the range is "all real numbers greater than 3 makes me think of 3 added to such a fraction. And since it must be greater than 3 that fraction should always give positive values.

So my first thought would be \(\displaystyle f(x)= 3+ \frac{1}{(x+ 4)^2}\). The numerator, 1, is positive, and the denominator is a square so the fraction is always positive.

The other two conditions are that the function increases for x< -4 and decreases for x> -4. What corrections, if any, do you need to make to the function? What must be true of a function where it is increasing or decreasing?
 
Determine the equation of a non-piecewise function that satisfies the following conditions.
-Increases over the domain (x<-4)
-decreases over the domain (x>-4)
D=(xER / x(cannot equal) -4)
R=(yER/ y>3)
I will guess that, by "E", you mean "is an element of"; by "D", you mean "the function's domain"; by the "R" in the last line, you mean "the function's range"; and, by the other R's, you mean "the real numbers".

What sorts of functions have you seen that have points excluded from their domains, without being piecewise? What functions have you seen that have restrictions on how high or how low they go (so, perhaps, they have horizontal asymptotes)? What sorts of functions have you studied recently?

Please reply showing all of your work and reasoning so far. Thank you! ;)
 
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