Prompt: S = x^3 + 3^x. Is S > 0
Notice that 3^x by itself is > 0 for all real values of x. By contrast, x^3 > 0 when x > 0, and x^3 < 0 when x < 0. When x > 0, both (3^x) and (x^3) are positive, so S must be positive.
This question is about the sum, so it's implicitly about how fast exponential function (e.g. 3^x) change with respect to power functions (x^3). This is a very advanced idea for GMAT math. In general, when exponential functions take off to infinity, they wildly outrun power functions.
Statement #1: x < 0
When x < 0, we know (x^3) is negative and (3^x) is positive. Let's try some values.
x = -1 --> S = (-1)^3 + 3^(-1) = -1 + 1/3 = -2/3 < 0
x = -2 --> S = (-2)^3 + 3^(-2) = -8 + 1/9 < 0
For these first two values of x, S is negative. As x gets more and more negative, (x^3) will be negative numbers of increasing magnitude, while (3^x) will be positive fractions getting closer and closer to zero, so S will get more and more negative.
That takes care of big negative values, but what happens we x is negative but between 0 and -1. Let's try some numbers there
x = -0.5 --> S = (-0.5)^3 + 3^(-0.5) = -1/8 + 1/sqrt(3)
sqrt(3) < 8, so 1/sqrt(3) > 1/8, so S is positive.
So, we can get negative values of x that make S either positive or negative, so Statement #1 is insufficient.
Statement #2: |x| > 1
Well, if x = anything positive, say x = 3, then S is positive. As we saw, if x = -2, S is negative. Knowing x has an absolute value greater than one, by itself, tells us nothing. Statement #2 is insufficient.
Statements #1 & #2 combined:
Now, we know x < 0 and |x| > 1, which means x < -1, in other words, x is a negative number with an absolute value greater than -1. Back in the argument under statement #1, we already saw that S is negative when x = -1, x = -2, and all larger negative numbers.
A more formal argument. When x < -1, we know
(a) (x^3) < -1 ---- (x^3) is a function that gets more & more negative as x gets larger.
(b) 0 < (3^x) < 1/3 ---- (3^x) will be a positive fraction of magnitude 1/3 or less.
S = big negative + small positive = negative. S must be negative when x < -1. Combined, the statements are sufficient.
Answer = C
Please let me know if you have any question on this.
Mike
