Inflection Points Of F(x) = X^2e^(13x) Calculation And Explanation

Jul 26, 2025 by ADMIN 67 views

Finding Inflection Points of f(x) = x^2e^(13x)

Hey guys! Today, we're diving into a fun calculus problem where we need to find the inflection points of the function f(x) = x²e^(13x). This might sound intimidating, but trust me, we'll break it down step by step. Inflection points are crucial in understanding the behavior of a function, as they indicate where the concavity changes – think of it as the points where the curve goes from being 'cup-up' to 'cup-down' or vice-versa. So, let's roll up our sleeves and get to it!

Understanding Inflection Points

First, let's clarify what inflection points are. In the simplest terms, an inflection point is a point on a curve at which the concavity changes. Concavity, in calculus terms, refers to the direction in which a curve bends. If a curve is concave up, it looks like a smile; if it's concave down, it looks like a frown. The inflection points are where these 'smiles' turn into 'frowns' and vice versa.

To find these points mathematically, we need to delve into the second derivative of the function. The second derivative tells us about the rate of change of the slope of the curve. If the second derivative is positive, the function is concave up; if it's negative, the function is concave down. At an inflection point, the second derivative will either be zero or undefined, and it will change sign around that point. This change in sign is key. A point where the second derivative is zero doesn't automatically qualify as an inflection point; we need to confirm that the concavity actually changes.

Now, why are inflection points so important? They give us valuable insights into the function's behavior. They help us sketch the graph accurately, identify intervals where the function is increasing or decreasing at an accelerating or decelerating rate, and understand the overall shape of the curve. In practical applications, inflection points can represent significant changes in a system – think of a population growth model where the inflection point might indicate the point of maximum growth rate before it starts to slow down due to limiting factors.

Calculating the First Derivative

Alright, let's jump into the math! To find the inflection points of f(x) = x²e^(13x), our first step is to calculate the first derivative, f'(x). Remember, the derivative gives us the slope of the tangent line at any point on the curve, which is crucial for understanding the function's rate of change.

For this, we'll need to use the product rule. The product rule states that if we have a function that is the product of two other functions, say u(x) and v(x), then the derivative of the product is given by:

(uv)' = u'v + uv'

In our case, we can consider u(x) = x² and v(x) = e^(13x). Let's find their derivatives separately:

u'(x) = d/dx (x²) = 2x
To find v'(x), we need the chain rule. The chain rule says that if we have a composite function, like e^(13x), we first differentiate the outer function (e^x) and then multiply by the derivative of the inner function (13x). So: v'(x) = d/dx (e^(13x)) = e^(13x) * d/dx (13x) = 13e^(13x)

Now we can apply the product rule:

f'(x) = u'v + uv' = (2x)(e^(13x)) + (x²)(13e^(13x))

Let's simplify this expression. We can factor out a common term of xe^(13x):

f'(x) = xe^(13x) (2 + 13x)

Great! We've got our first derivative. This expression tells us how the function is increasing or decreasing at any given point. But to find inflection points, we need the second derivative, so let's move on to that.

Finding the Second Derivative

Okay, guys, we're making progress! We've got the first derivative, f'(x) = xe^(13x)(2 + 13x). Now, to find those inflection points, we need to calculate the second derivative, f''(x). This will tell us about the concavity of our function.

Looking at f'(x), we see that we'll need to use the product rule again, but this time it might look a bit more complex. We have a product of three terms: x, e^(13x), and (2 + 13x). To simplify this, let's group the first two terms together and consider them as one function, and the third term as another function. So, we can think of f'(x) as a product of:

u(x) = xe^(13x)
v(x) = (2 + 13x)

We already know the derivative of v(x), which is straightforward:

v'(x) = d/dx (2 + 13x) = 13

Now, let's find the derivative of u(x). This itself requires the product rule! We can break u(x) down further into:

p(x) = x
q(x) = e^(13x)

We know:

p'(x) = d/dx (x) = 1
q'(x) = d/dx (e^(13x)) = 13e^(13x) (from our previous calculations)

Applying the product rule to u(x):

u'(x) = p'q + pq' = (1)(e^(13x)) + (x)(13e^(13x)) = e^(13x) + 13xe^(13x)

We can factor out e^(13x) to get:

u'(x) = e^(13x)(1 + 13x)

Okay, we've got u'(x). Now we can finally find f''(x) using the product rule on u(x) and v(x):

f''(x) = u'v + uv' = [e^(13x)(1 + 13x)](2 + 13x) + xe^(13x)

This looks a bit messy, but let's simplify it. We can factor out e^(13x):

f''(x) = e^(13x) [(1 + 13x)(2 + 13x) + 13x]

Now let's expand the terms inside the brackets:

f''(x) = e^(13x) [2 + 13x + 26x + 169x² + 13x]

Combine like terms:

f''(x) = e^(13x) [169x² + 52x + 2]

Woohoo! We've found the second derivative. This expression is crucial for finding the inflection points. Next, we'll set this equal to zero and solve for x.

Finding Potential Inflection Points

Alright, team, we've got the second derivative: f''(x) = e^(13x) [169x² + 52x + 2]. Now, the next step in our quest for inflection points is to find the values of x where f''(x) = 0. These points are our potential inflection points, as they are where the concavity might change.

Notice that f''(x) is a product of two terms: e^(13x) and [169x² + 52x + 2]. Let's consider each term separately.

First, e^(13x). The exponential function e^u is always positive for any real number u. Therefore, e^(13x) will never be zero. This means we can ignore this term when solving for f''(x) = 0.

Now, let's focus on the quadratic expression: 169x² + 52x + 2. To find where this is equal to zero, we need to solve the quadratic equation:

169x² + 52x + 2 = 0

This doesn't look like it factors easily, so we'll use the quadratic formula. Remember the quadratic formula? For an equation of the form ax² + bx + c = 0, the solutions are given by:

x = [-b ± √(b² - 4ac)] / (2a)

In our case, a = 169, b = 52, and c = 2. Let's plug these values into the formula:

x = [-52 ± √(52² - 4 * 169 * 2)] / (2 * 169)

Let's simplify this step by step. First, calculate the discriminant (the part under the square root):

b² - 4ac = 52² - 4 * 169 * 2 = 2704 - 1352 = 1352

So, our equation becomes:

x = [-52 ± √1352] / 338

We can simplify the square root a bit. 1352 can be factored as 4 * 338, so √1352 = √(4 * 338) = 2√338. Thus:

x = [-52 ± 2√338] / 338

We can simplify this fraction by dividing both the numerator and denominator by 2:

x = [-26 ± √338] / 169

So, we have two potential inflection points:

x₁ = (-26 - √338) / 169
x₂ = (-26 + √338) / 169

These are the x-values where the second derivative is zero. But remember, we're not done yet! We need to check if the concavity actually changes at these points.

Verifying Inflection Points and Determining C and D

Okay, rockstars, we've found the potential inflection points: x₁ = (-26 - √338) / 169 and x₂ = (-26 + √338) / 169. But, as we discussed earlier, just because the second derivative is zero at these points doesn't automatically make them inflection points. We need to verify that the concavity of the function actually changes at these points.

To do this, we'll use the second derivative test. This involves choosing test values in the intervals created by our potential inflection points and plugging them into the second derivative, f''(x) = e^(13x) [169x² + 52x + 2]. The sign of f''(x) in each interval will tell us about the concavity of the function in that interval.

Our potential inflection points divide the number line into three intervals:

x < (-26 - √338) / 169
(-26 - √338) / 169 < x < (-26 + √338) / 169
x > (-26 + √338) / 169

Let's pick test values in each interval. To make things easier, let's approximate our potential inflection points:

x₁ = (-26 - √338) / 169 ≈ (-26 - 18.38) / 169 ≈ -0.26
x₂ = (-26 + √338) / 169 ≈ (-26 + 18.38) / 169 ≈ -0.045

So, our intervals are approximately: x < -0.26, -0.26 < x < -0.045, and x > -0.045.

Now, let's choose test values:

For x < -0.26, let's pick x = -1
For -0.26 < x < -0.045, let's pick x = -0.1
For x > -0.045, let's pick x = 0

Let's plug these values into f''(x) = e^(13x) [169x² + 52x + 2]. Remember, we only care about the sign of f''(x), since e^(13x) is always positive.

f''(-1) = e^(-13) [169(-1)² + 52(-1) + 2] = e^(-13) [169 - 52 + 2] = e^(-13) [119] > 0 (Concave Up)
f''(-0.1) = e^(-1.3) [169(-0.1)² + 52(-0.1) + 2] = e^(-1.3) [1.69 - 5.2 + 2] = e^(-1.3) [-1.51] < 0 (Concave Down)
f''(0) = e^(0) [169(0)² + 52(0) + 2] = 1 [2] = 2 > 0 (Concave Up)

We can see that the concavity changes at both x₁ and x₂. The function goes from concave up to concave down at x₁, and from concave down to concave up at x₂. Therefore, both points are indeed inflection points!

The problem states that C < D, so:

C = (-26 - √338) / 169
D = (-26 + √338) / 169

Conclusion

And there you have it, folks! We've successfully found the inflection points of the function f(x) = x²e^(13x). We navigated through the product rule, the chain rule, the quadratic formula, and the second derivative test. It was a journey, but we got there!

We determined that the inflection points occur at x = C = (-26 - √338) / 169 and x = D = (-26 + √338) / 169. These points are crucial for understanding the shape and behavior of the function. Remember, inflection points are where the concavity changes, providing valuable insights into how the function is curving.

I hope this breakdown was helpful and clear. Keep practicing these calculus concepts, and you'll become a pro in no time. Keep up the great work, guys, and happy calculating!