Finding P: Arithmetic Sequence Of Derivatives And Function

Hey guys! Let's dive into a cool math problem where we're given a function, its derivatives, and an arithmetic sequence. We need to find the value of a constant, p. This might sound intimidating, but trust me, we'll break it down step by step. So, grab your thinking caps, and let's get started!

Understanding the Problem

Okay, so the problem states that we have a function $f(x) = x^3 + 3x^2 + p$. The second derivative of the function at $x=2$ ($f''(2)$), the first derivative at $x=2$ ($f'(2)$), and the function itself at $x=2$ ($f(2)$) form an arithmetic sequence. Our mission, should we choose to accept it, is to find the value of p. To effectively tackle this problem, let's first clearly define the key concepts involved. We have a polynomial function and its derivatives, and we are also dealing with the properties of an arithmetic sequence. These are the building blocks we need to solve this.

The Function and Its Derivatives

The function we're given is a polynomial, specifically a cubic function. Polynomial functions are those that involve only non-negative integer powers of the variable (in this case, x). To solve the problem, we'll need to find the first and second derivatives of this function. Remember, the derivative of a function gives us the slope of the tangent line at any point on the function's graph. It tells us how the function is changing. So, finding the derivatives is crucial for understanding the function's behavior.

To find the first derivative, $f'(x)$, we'll use the power rule of differentiation. The power rule states that if $f(x) = ax^n$, then $f'(x) = nax^{n-1}$. Applying this rule to each term in our function, we can determine the first derivative. Next, we'll need to find the second derivative, $f''(x)$. The second derivative tells us about the concavity of the function – whether it's curving upwards or downwards. To find the second derivative, we simply differentiate the first derivative again, using the same power rule. This process will give us the expressions we need to evaluate at $x = 2$.

Arithmetic Sequences

Now, let's talk about arithmetic sequences. An arithmetic sequence is a sequence of numbers where the difference between any two consecutive terms is constant. This constant difference is called the common difference. For example, 2, 4, 6, 8... is an arithmetic sequence with a common difference of 2. The key property of an arithmetic sequence that we'll use is that the middle term is the average of the first and third terms. This property will allow us to set up an equation and solve for p. Understanding this fundamental characteristic of arithmetic sequences is crucial for linking the derivatives and function values together and ultimately finding the solution.

Calculating the Derivatives

Alright, let's get our hands dirty with some calculus! We need to find the first and second derivatives of our function, $f(x) = x^3 + 3x^2 + p$. Remember, the power rule is our best friend here. It's all about bringing down the exponent and reducing it by one.

Finding the First Derivative, f'(x)

Let's start with the first derivative, $f'(x)$. We'll differentiate each term of the function separately. The derivative of $x^3$ is $3x^2$ (3 comes down, and the exponent reduces from 3 to 2). The derivative of $3x^2$ is $6x$ (2 comes down and multiplies with 3, and the exponent reduces from 2 to 1). And the derivative of the constant p is 0 (the derivative of a constant is always zero). So, putting it all together, we have:

$f'(x) = 3x^2 + 6x$

See? Not too scary, right? The power rule makes differentiation quite manageable. Now we have the expression for the first derivative, which will be important when we evaluate it at $x = 2$. This gives us the rate of change of the function at any given point, which is essential for understanding its behavior.

Finding the Second Derivative, f''(x)

Next up, let's find the second derivative, $f''(x)$. Remember, the second derivative is simply the derivative of the first derivative. So, we'll differentiate $f'(x) = 3x^2 + 6x$. Again, we'll use the power rule. The derivative of $3x^2$ is $6x$, and the derivative of $6x$ is 6. Therefore:

$f''(x) = 6x + 6$

Awesome! We've found both the first and second derivatives. These expressions are the foundation for the next step, where we'll evaluate them at $x = 2$ and use the arithmetic sequence property to solve for p. We're making good progress, guys! The key is to break down the problem into manageable steps, and we're doing just that.

Evaluating at x = 2

Now that we have the expressions for $f(x)$, $f'(x)$, and $f''(x)$, it's time to plug in $x = 2$. This will give us the specific values of the function and its derivatives at that point, which we need to use the arithmetic sequence information.

Calculating f(2)

First, let's find $f(2)$. We substitute $x = 2$ into the original function: $f(x) = x^3 + 3x^2 + p$.

$f(2) = (2)^3 + 3(2)^2 + p = 8 + 12 + p = 20 + p$

So, $f(2) = 20 + p$. This is one of the terms in our arithmetic sequence, and it involves the unknown p, which is what we're trying to find. Keep in mind that this is the value of the function itself at $x=2$, and it's directly linked to the constant p.

Calculating f'(2)

Next, let's calculate $f'(2)$. We substitute $x = 2$ into the first derivative: $f'(x) = 3x^2 + 6x$.

$f'(2) = 3(2)^2 + 6(2) = 3(4) + 12 = 12 + 12 = 24$

So, $f'(2) = 24$. This is the value of the first derivative at $x = 2$, which represents the slope of the tangent line to the function at that point. This value is a fixed number and doesn't involve p.

Calculating f''(2)

Finally, let's calculate $f''(2)$. We substitute $x = 2$ into the second derivative: $f''(x) = 6x + 6$.

$f''(2) = 6(2) + 6 = 12 + 6 = 18$

So, $f''(2) = 18$. This is the value of the second derivative at $x = 2$, which tells us about the concavity of the function at that point. Like $f'(2)$, this value is also a fixed number.

Now we have all three values: $f(2) = 20 + p$, $f'(2) = 24$, and $f''(2) = 18$. These three values form our arithmetic sequence, and we're ready to use that information to solve for p. We're on the home stretch, guys!

Using the Arithmetic Sequence Property

We've calculated $f''(2) = 18$, $f'(2) = 24$, and $f(2) = 20 + p$. The problem tells us that these values form an arithmetic sequence. Remember, the key property of an arithmetic sequence is that the middle term is the average of the first and third terms. This is our golden ticket to solving for p!

Setting up the Equation

In our sequence, $f'(2)$ is the middle term. So, it must be the average of $f''(2)$ and $f(2)$. This gives us the following equation:

f'(2) = rac{f''(2) + f(2)}{2}

Now we can substitute the values we calculated earlier:

24 = rac{18 + (20 + p)}{2}

This equation relates all the values we've found, and most importantly, it includes our unknown, p. Now it's just a matter of solving this equation for p. We've transformed a calculus problem into a simple algebraic equation, which is pretty cool!

Solving for p

Let's solve the equation for p. First, we can multiply both sides of the equation by 2 to get rid of the fraction:

$2 * 24 = 18 + (20 + p)$

$48 = 18 + 20 + p$

Now, we can simplify the right side:

$48 = 38 + p$

Finally, to isolate p, we subtract 38 from both sides:

$48 - 38 = p$

$10 = p$

So, we've found it! The value of p is 10. We've successfully navigated through the calculus and the arithmetic sequence property to find the solution. Give yourselves a pat on the back, guys!

The Answer

Therefore, the value of $p$ is 10. We started with a function, found its derivatives, used the properties of arithmetic sequences, and solved for an unknown constant. This problem shows how different mathematical concepts can come together to solve a single problem. You guys did great! Remember, the key to tackling complex problems is to break them down into smaller, manageable steps. And that's exactly what we did here. Keep practicing, and you'll become math ninjas in no time!