Re: If g(x) = 2^x + x, how many solutions satisfy the equation g(x) = 2?
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Updated on: 07 Feb 2016, 20:49
Usually when I see problems like this (graph, number of times a value is hit, exponential equation, etc.), I try to find a key inflection x value. For any a^b question, the inflection x will often be 0. Be careful with this rule of thumb, of course. Don't trust it blindly. Once we've defined our inflection point (x = 0 for this question), we break our equation into three parts: x = 0, x < 0, x > 0. We want to know what the function is doing in each of these ranges. For this specific question, we want to know if each part is upward-trending or downward-trending.
x = 0:
2^0 + 0 = 1. g(1) = 1. Next question: are both x < 0 and x > 0 upward trending? Downward trending? Etc.
x < 0:
\(2^x + x, x < 0\)
For simplicity, let's flip the signs on everything so we're not dealing with awkward negatives to throw us off: \(\frac{1}{{2^x}} - x, x > 0\). This means that, returning to x < 0, the lesser x is, the more negative y is. The only unsurety is for -1 < x < 0. If you know calculus and limits, \(\frac{1}{{2^x}} + x\) as x approaches 0 will become 1/1 + 0, which is 1. We conclude that the trend is strictly downwards.
x > 0:
\(2^x + x, x > 0\)
\(2^x\) is a hyperbola when x > 0. We know that it will pass through all positive x points and, for x > 0, all y points above 1. Our trend is strictly upwards.
Since g(1) = 1, g(x<0) < 1, g(x>0) > 1, x>0 is the upward trending half of a hyperbola, and x>0 will cover the case of g(1) = 1, we know that there will be only one point where g(x) crosses 2.
Edit: Added a warning about x = 0 as an inflection point.
Originally posted by
Beixi88 on 07 Feb 2016, 20:33.
Last edited by
Beixi88 on 07 Feb 2016, 20:49, edited 2 times in total.