1. Given that $$ y=x(x+1)^{2} $$, calarlate the mancimum value of $$ y $$ ?

Question

UpStudy AI · Accepted Answer

To find the maximum value of the function $$ y = x(x+1)^{2} $$, we will follow these steps:

1. **Differentiate the function** to find the critical points.
2. **Set the derivative equal to zero** to find the values of $$ x $$ that could give us maximum or minimum points.
3. **Evaluate the second derivative** to determine whether these points are maxima or minima.
4. **Calculate the maximum value of $$ y $$** at the critical points.

Let's start by differentiating the function $$ y $$.

### Step 1: Differentiate the function

The function is given by:
$$
y = x(x+1)^{2}
$$

Using the product rule, we differentiate $$ y $$:
$$
\frac{dy}{dx} = (x+1)^{2} + x \cdot 2(x+1) \cdot \frac{d}{dx}(x+1)
$$
$$
= (x+1)^{2} + 2x(x+1)
$$

Now, let's simplify this expression.

### Step 2: Set the derivative equal to zero

We will set the derivative equal to zero and solve for $$ x $$:
$$
\frac{dy}{dx} = (x+1)^{2} + 2x(x+1) = 0
$$

Now, let's simplify and solve this equation.

### Step 3: Solve for critical points

Let's simplify the derivative:
$$
\frac{dy}{dx} = (x+1)^{2} + 2x(x+1) = (x+1)(x+1 + 2x) = (x+1)(3x + 1)
$$

Setting this equal to zero gives us:
$$
(x+1)(3x + 1) = 0
$$

This results in two equations:
1. $$ x + 1 = 0 $$  →  $$ x = -1 $$
2. $$ 3x + 1 = 0 $$  →  $$ x = -\frac{1}{3} $$

### Step 4: Determine the nature of the critical points

Next, we will evaluate the second derivative to determine if these points are maxima or minima.

Let's calculate the second derivative $$ \frac{d^2y}{dx^2} $$.

### Step 5: Calculate the second derivative

We will differentiate $$ \frac{dy}{dx} = (x+1)(3x + 1) $$ again.

Using the product rule:
$$
\frac{d^2y}{dx^2} = (3x + 1) + (x + 1) \cdot 3
$$
$$
= 3x + 1 + 3(x + 1) = 3x + 1 + 3x + 3 = 6x + 4
$$

Now, we will evaluate the second derivative at the critical points $$ x = -1 $$ and $$ x = -\frac{1}{3} $$.

1. For $$ x = -1 $$:
$$
\frac{d^2y}{dx^2} = 6(-1) + 4 = -6 + 4 = -2 \quad (\text{maxima})
$$

2. For $$ x = -\frac{1}{3} $$:
$$
\frac{d^2y}{dx^2} = 6\left(-\frac{1}{3}\right) + 4 = -2 + 4 = 2 \quad (\text{minima})
$$

### Step 6: Calculate the maximum value of $$ y $$

Now, we will calculate the value of $$ y $$ at the maximum point $$ x = -1 $$:
$$
y = (-1)((-1)+1)^{2} = (-1)(0) = 0
$$

Thus, the maximum value of $$ y $$ is:
$$
\boxed{0}
$$
The maximum value of $$ y $$ is 0.

Given that , calarlate the mancimum
value of ?

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