Find the value of x at the minimum point of the curve y = x3 + x2 - x + 1

To find the minimum point of the curve $y = x^3 + x^2 - x + 1$, we follow the principles of calculus used in finding stationary points.

1. Analyze the question

We are looking for the value of $x$ where the function reaches its minimum. This is a stationary point problem where we must find the derivative, set it to zero, and use the second derivative test to confirm if it is a minimum.

2. Explain the concept

• First Derivative ($\frac{dy}{dx}$): This represents the slope of the curve. At any stationary point (maximum or minimum), the slope is zero.
• Second Derivative ($\frac{d^2y}{dx^2}$): This helps us classify the stationary point. • If $\frac{d^2y}{dx^2} > 0$, the point is a minimum. • If $\frac{d^2y}{dx^2} < 0$, the point is a maximum.

3. Show the solution

Step 1: Find the first derivative. Given $y = x^3 + x^2 - x + 1$ $\frac{dy}{dx} = 3x^2 + 2x - 1$

Step 2: Set the derivative to zero to find stationary points. $3x^2 + 2x - 1 = 0$

Factorize the quadratic equation: $3x^2 + 3x - x - 1 = 0$ $3x(x + 1) - 1(x + 1) = 0$ $(3x - 1)(x + 1) = 0$

So, $x = \frac{1}{3}$ or $x = -1$.

Step 3: Find the second derivative to classify the points. $\frac{d^2y}{dx^2} = \frac{d}{dx}(3x^2 + 2x - 1) = 6x + 2$

Step 4: Test the points.

• For $x = -1$: $\frac{d^2y}{dx^2} = 6(-1) + 2 = -4$. Since $-4 < 0$, this is a maximum.
• For $x = \frac{1}{3}$: $\frac{d^2y}{dx^2} = 6(\frac{1}{3}) + 2 = 2 + 2 = 4$. Since $4 > 0$, this is a minimum.

4. Identify the answer

The value of $x$ at the minimum point is $\frac{1}{3}$. This matches option A.

5. Tip

When dealing with cubic functions ($ax^3 + bx^2 + cx + d$), the stationary points are always found by solving the quadratic equation resulting from the first derivative. Always remember: "Positive second derivative = Minimum (think 'positive/happy face'), Negative second derivative = Maximum (think 'negative/sad face')."