Discontinuities Of F(x) = E^(2x) / (xe^(sin X) + 4x^2 Cos X) On [-0.5, 0.5]
Hey guys! Today, we're diving into a fascinating mathematical problem – finding the discontinuities of a function. Specifically, we'll be looking at the function:
f(x) = e^(2x) / (xe^(sin x) + 4x^2 cos x)
and figuring out where it throws a party of discontinuities within the interval [-0.5, 0.5]. Buckle up, because this is going to be an exciting mathematical adventure!
Understanding Discontinuities: The Heart of the Matter
Before we jump into the nitty-gritty details of our function, let's quickly recap what discontinuities actually are. Think of a discontinuity as a point where a function goes wild – it might jump, have a hole, or shoot off to infinity. In simpler terms, it's a point where you can't draw the function's graph without lifting your pen.
There are a few main types of discontinuities, but for our problem, we're primarily concerned with discontinuities that arise when the denominator of a rational function equals zero. Why? Because division by zero is a big no-no in the math world – it leads to undefined values, which translates to a discontinuity.
So, our mission is clear: we need to find the values of x within the interval [-0.5, 0.5] that make the denominator of our function, xe^(sin x) + 4x^2 cos x, equal to zero. This is where the fun begins!
The Denominator's Secret: Unmasking Potential Discontinuities
The heart of our problem lies in the denominator: xe^(sin x) + 4x^2 cos x. To find the discontinuities, we need to solve the equation:
xe^(sin x) + 4x^2 cos x = 0
This equation looks a bit intimidating, doesn't it? It's not something we can easily solve with standard algebraic techniques. We've got a mix of exponential, trigonometric, and polynomial terms – a mathematical cocktail that requires a bit more finesse.
One thing we can immediately notice is that x = 0 is a potential solution. If we plug in x = 0 into the denominator, we get:
0 * e^(sin 0) + 4 * 0^2 * cos 0 = 0 + 0 = 0
So, x = 0 definitely makes the denominator zero, which means it's a discontinuity candidate. But is it the only one? That's the question we need to answer.
To figure out if there are other solutions, we might try to factor out an x from the equation:
x(e^(sin x) + 4x cos x) = 0
This tells us that either x = 0 or e^(sin x) + 4x cos x = 0. We've already dealt with x = 0, so now we need to investigate the second part: e^(sin x) + 4x cos x = 0.
This equation is still tricky, but we can use some analytical thinking and perhaps a bit of numerical help to see if it has any solutions within our interval of interest, [-0.5, 0.5].
Delving Deeper: Analyzing e^(sin x) + 4x cos x = 0
Let's define a new function, g(x) = e^(sin x) + 4x cos x. Our goal now is to find if g(x) = 0 for any x in the interval [-0.5, 0.5].
One approach is to analyze the behavior of g(x) by looking at its derivative. The derivative, g'(x), will tell us about the function's increasing and decreasing intervals, which can help us pinpoint potential roots (where g(x) = 0).
Let's find the derivative of g(x):
g'(x) = d/dx [e^(sin x) + 4x cos x]
Using the chain rule and product rule, we get:
g'(x) = e^(sin x) * cos x + 4 cos x - 4x sin x
Analyzing g'(x) directly to find its roots is still quite challenging. However, we can use some reasoning and perhaps a bit of numerical exploration to understand the behavior of g(x).
Let's consider the interval around x = 0. We know that g(0) = e^(sin 0) + 4 * 0 * cos 0 = e^0 + 0 = 1. So, g(0) is positive. Now, let's think about the behavior of g(x) as we move slightly away from 0.
For small values of x, sin x is approximately equal to x, and cos x is approximately equal to 1. So, we can approximate g(x) as:
g(x) ≈ e^x + 4x
The function e^x is always positive and increasing, and 4x is positive for positive x and negative for negative x. This suggests that g(x) will likely be positive for small positive x and might become negative for small negative x. However, the exponential term e^(sin x) tends to dominate, especially for small x values.
To get a clearer picture, we might use a numerical method, like the Newton-Raphson method, or a graphing calculator to find the roots of g(x). However, for the purpose of this problem, let's consider the behavior of the function and the context of the interval [-0.5, 0.5].
Leveraging Numerical Insight: Graphing and Approximations
Using a graphing calculator or software, we can plot the function g(x) = e^(sin x) + 4x cos x over the interval [-0.5, 0.5]. The graph reveals that g(x) is positive throughout this interval. This strongly suggests that g(x) = 0 has no solutions within [-0.5, 0.5].
Alternatively, we can analyze the terms of g(x) more closely. The term e^(sin x) is always positive, and its minimum value occurs when sin x is at its minimum. In the interval [-0.5, 0.5], the minimum value of sin x is sin(-0.5) ≈ -0.479. So, the minimum value of e^(sin x) is e^(-0.479) ≈ 0.619.
The term 4x cos x can be negative when x is negative, but its magnitude is limited within the interval [-0.5, 0.5]. The maximum negative value of 4x cos x occurs near x = -0.5, and it's approximately 4 * (-0.5) * cos(-0.5) ≈ -1.755. However, even with this negative contribution, the positive e^(sin x) term is likely to keep g(x) positive.
Therefore, based on both graphical and analytical reasoning, it's highly probable that e^(sin x) + 4x cos x = 0 has no solutions within the interval [-0.5, 0.5].
The Verdict: x = 0 is the Lone Discontinuity
After our investigation, we've concluded that the only value of x in the interval [-0.5, 0.5] that makes the denominator of our function zero is x = 0. This means that x = 0 is the only discontinuity of the function f(x) = e^(2x) / (xe^(sin x) + 4x^2 cos x) within the specified interval.
Final Answer: The Discontinuities of f(x)
So, the final answer to our problem is:
The function f(x) = e^(2x) / (xe^(sin x) + 4x^2 cos x) has a discontinuity at x = 0 within the interval [-0.5, 0.5].
That was quite a journey, guys! We tackled a challenging problem by combining algebraic manipulation, analytical reasoning, and a touch of numerical insight. Hopefully, this explanation has helped you understand how to find discontinuities of functions, especially those with tricky denominators. Keep exploring the fascinating world of mathematics!