Concave Down Intervals For F(x) = (4x^2 - 2x)e^{-2x}

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Hey guys, welcome back to Plastik Magazine! Today, we're diving deep into the fascinating world of calculus to tackle a problem that's all about understanding the shape of a function's graph. Specifically, we're going to find all the intervals on which the graph of $f(x)=\left(4 x^2-2 x\right) e^{-2 x}$ is concave down. This is a classic problem that requires us to use the second derivative of the function to determine its concavity. So, grab your calculators, a fresh cup of coffee, and let's get this done!

Understanding Concavity: What It Means for a Graph

Before we jump into the calculations, let's quickly recap what concavity means in calculus, guys. When we talk about a function's graph being concave down, we're essentially describing a shape that looks like an upside-down bowl or a frown. Mathematically, a function $f(x)$ is concave down on an interval if its second derivative, $f''(x)$, is negative on that interval. Conversely, if $f''(x)$ is positive, the graph is concave up (like a right-side-up bowl). The points where the concavity changes are called inflection points, and they occur where $f''(x) = 0$ or where $f''(x)$ is undefined. To find the intervals of concavity, we'll need to:

1. Find the first derivative ($f'(x)$) of the given function.
2. Find the second derivative ($f''(x)$) by differentiating $f'(x)$.
3. Determine where $f''(x) < 0$. This will give us the intervals where the function is concave down.

Our function for today is $f(x)=\left(4 x^2-2 x\right) e^{-2 x}$. This function is a product of a quadratic term and an exponential term. Both of these are differentiable everywhere, which simplifies things for us – we don't need to worry about undefined points in our derivatives.

Step 1: Calculating the First Derivative ($f'(x)$)

Alright team, the first major step is to find the first derivative of our function $f(x)=\left(4 x^2-2 x\right) e^{-2 x}$. Since this is a product of two functions, we'll need to use the product rule. Remember the product rule, guys? If $h(x) = u(x)v(x)$, then $h'(x) = u'(x)v(x) + u(x)v'(x)$.

Let $u(x) = 4x^2 - 2x$ and $v(x) = e^{-2x}$.

Now, let's find their derivatives:

• $u'(x) = \frac{d}{dx}(4x^2 - 2x) = 8x - 2$
• $v'(x) = \frac{d}{dx}(e^{-2x})$. Using the chain rule here, the derivative of $e^w$ is $e^w$, and the derivative of $-2x$ is $-2$. So, $v'(x) = -2e^{-2x}$.

Now, let's plug these into the product rule formula:

$f'(x) = u'(x)v(x) + u(x)v'(x)$

$f'(x) = (8x - 2)e^{-2x} + (4x^2 - 2x)(-2e^{-2x})$

We can factor out $e^{-2x}$ to simplify:

$f'(x) = e^{-2x} [ (8x - 2) - 2(4x^2 - 2x) ]$

Let's distribute the $-2$ inside the brackets:

$f'(x) = e^{-2x} [ 8x - 2 - 8x^2 + 4x ]$

Combine like terms inside the brackets:

$f'(x) = e^{-2x} [ -8x^2 + 12x - 2 ]$

And we can factor out a $-2$ from the quadratic part to make it even cleaner:

$f'(x) = -2e^{-2x} [ 4x^2 - 6x + 1 ]$

So, there you have it – the first derivative! This tells us about the slope of the original function's graph. Now, we need to take it a step further to understand its shape.

Step 2: Calculating the Second Derivative ($f''(x)$)

Alright, smart cookies, let's move on to the main event: finding the second derivative, $f''(x)$. This is crucial because the sign of the second derivative tells us about the concavity. We'll be differentiating our $f'(x) = -2e^{-2x} (4x^2 - 6x + 1)$. Again, we have a product of two functions: $-2e^{-2x}$ and $(4x^2 - 6x + 1)$. We'll need the product rule once more.

Let $p(x) = -2e^{-2x}$ and $q(x) = 4x^2 - 6x + 1$.

Let's find their derivatives:

• $p'(x) = \frac{d}{dx}(-2e^{-2x})$. Using the chain rule, this is $-2 \times (-2e^{-2x}) = 4e^{-2x}$.
• $q'(x) = \frac{d}{dx}(4x^2 - 6x + 1) = 8x - 6$.

Now, applying the product rule for $f''(x) = p'(x)q(x) + p(x)q'(x)$:

$f''(x) = (4e^{-2x})(4x^2 - 6x + 1) + (-2e^{-2x})(8x - 6)$

Let's factor out $e^{-2x}$ to simplify the expression:

$f''(x) = e^{-2x} [ 4(4x^2 - 6x + 1) - 2(8x - 6) ]$

Now, distribute the constants inside the brackets:

$f''(x) = e^{-2x} [ (16x^2 - 24x + 4) - (16x - 12) ]$

Be careful with the signs when distributing the $-2$!

$f''(x) = e^{-2x} [ 16x^2 - 24x + 4 - 16x + 12 ]$

Combine like terms inside the brackets:

$f''(x) = e^{-2x} [ 16x^2 - 40x + 16 ]$

We can factor out a $16$ from the quadratic expression to make it even tidier:

$f''(x) = 16e^{-2x} [ x^2 - \frac{40}{16}x + \frac{16}{16} ]$

Simplifying the fractions:

$f''(x) = 16e^{-2x} [ x^2 - \frac{5}{2}x + 1 ]$

And we can factor out a 16 from the quadratic expression for even more tidiness:

$f''(x) = 16e^{-2x} [ x^2 - \frac{5}{2}x + 1 ]$

Alternatively, we can factor out a $16$ from the entire expression:

$f''(x) = 16e^{-2x} \left(x^2 - \frac{5}{2}x + 1\right)$

To make dealing with fractions easier, let's factor out a $2$ from the parenthesis instead:

$f''(x) = 16e^{-2x} \times \frac{1}{2} \left(2x^2 - 5x + 2\right)$

$f''(x) = 8e^{-2x} \left(2x^2 - 5x + 2\right)$

This is our second derivative. This is what we'll use to find the intervals of concavity. Remember, we are looking for where $f''(x) < 0$.

Step 3: Finding Intervals Where $f''(x) < 0$

Now for the final stretch, guys! We need to find the intervals where $f''(x) < 0$. Our second derivative is $f''(x) = 8e^{-2x} (2x^2 - 5x + 2)$.

Let's analyze the components of this expression:

• The term $8e^{-2x}$ is always positive for all real values of $x$. This is because $e$ raised to any real power is always positive, and multiplying by $8$ doesn't change that.

• Therefore, the sign of $f''(x)$ is determined entirely by the sign of the quadratic factor: $(2x^2 - 5x + 2)$.

So, we need to find where $2x^2 - 5x + 2 < 0$.

To do this, let's first find the roots of the quadratic equation $2x^2 - 5x + 2 = 0$. We can use factoring or the quadratic formula. Let's try factoring:

We are looking for two numbers that multiply to $2 \times 2 = 4$ and add up to $-5$. These numbers are $-1$ and $-4$. So, we can rewrite the middle term:

$2x^2 - 4x - x + 2 = 0$

Now, factor by grouping:

$2x(x - 2) - 1(x - 2) = 0$

$(2x - 1)(x - 2) = 0$

This gives us our roots:

• $2x - 1 = 0 \implies x = \frac{1}{2}$
• $x - 2 = 0 \implies x = 2$

These roots, $x = \frac{1}{2}$ and $x = 2$, divide the number line into three intervals: $\left(-\infty, \frac{1}{2}\right)$, $\left(\frac{1}{2}, 2\right)$, and $\left(2, \infty\right)$. We need to test the sign of the quadratic $2x^2 - 5x + 2$ in each of these intervals.

Interval 1: $\left(-\infty, \frac{1}{2}\right)$

Let's pick a test value, say $x = 0$ (which is less than $\frac{1}{2}$).

Plug $x=0$ into $(2x - 1)(x - 2)$: $(2(0) - 1)(0 - 2) = (-1)(-2) = 2$. This is positive.

So, $f''(x) > 0$ on $\left(-\infty, \frac{1}{2}\right)$. The function is concave up here.

Interval 2: $\left(\frac{1}{2}, 2\right)$

Let's pick a test value, say $x = 1$ (which is between $\frac{1}{2}$ and $2$).

Plug $x=1$ into $(2x - 1)(x - 2)$: $(2(1) - 1)(1 - 2) = (1)(-1) = -1$. This is negative.

So, $f''(x) < 0$ on $\left(\frac{1}{2}, 2\right)$. This is where our function is concave down!

Interval 3: $\left(2, \infty\right)$

Let's pick a test value, say $x = 3$ (which is greater than $2$).

Plug $x=3$ into $(2x - 1)(x - 2)$: $(2(3) - 1)(3 - 2) = (5)(1) = 5$. This is positive.

So, $f''(x) > 0$ on $\left(2, \infty\right)$. The function is concave up here.

Conclusion: The Concave Down Intervals

Alright guys, we've done it! By analyzing the sign of the second derivative, we've found the intervals where the graph of $f(x)=\left(4 x^2-2 x\right) e^{-2 x}$ is concave down. Remember, concavity is determined by the sign of the second derivative, and concave down means $f''(x) < 0$.

We found that the sign of $f''(x)$ is determined by the quadratic $2x^2 - 5x + 2$. The roots of this quadratic are $x = \frac{1}{2}$ and $x = 2$. By testing values in the intervals created by these roots, we discovered that the quadratic is negative only between $\frac{1}{2}$ and $2$.

Therefore, the graph of $f(x)=\left(4 x^2-2 x\right) e^{-2 x}$ is concave down on the interval $\boxed{\left(\frac{1}{2}, 2\right)}$.

Keep practicing these types of problems, guys! Understanding concavity is a fundamental concept in calculus that helps us visualize and interpret the behavior of functions. If you have any questions or want to try another problem, drop a comment below. Until next time, stay curious and keep exploring the amazing world of mathematics!