Question

Find the stationary values of the following (check whether they are relative maxima or minima or inflection points), assuming the domain to be the interval \((0, \infty)\) $$(a) y=x^{3}-3 x+5$$ $$(b) y=\frac{1}{3} x^{3}-x^{2}+x+10$$ $$(c) y=-x^{3}+4.5 x^{2}-6 x+6$$

Short answer

(a) Local minimum at \(x=1\); (b) Inflection point at \(x=1\); (c) Local minimum at \(x=1\), local maximum at \(x=2\).

Step-by-step solution

Differentiate the Function (a)

First, we need to differentiate the function \(y = x^3 - 3x + 5\) with respect to \(x\). The derivative is \(\frac{dy}{dx} = 3x^2 - 3\).

Find Critical Points for (a)

To find the critical points, set the derivative from Step 1 equal to zero: \(3x^2 - 3 = 0\). Solving gives \(x^2 = 1\), so \(x = 1\) or \(x = -1\). Since we consider the interval \((0, \infty)\), only \(x = 1\) is valid.

Second Derivative Test for (a)

Find the second derivative: \(\frac{d^2y}{dx^2} = 6x\). At \(x = 1\), the second derivative is \(6\cdot1 = 6\), which is positive. This indicates a local minimum at \(x = 1\).

Differentiate the Function (b)

Differentiate \(y = \frac{1}{3}x^3 - x^2 + x + 10\) to get \(\frac{dy}{dx} = x^2 - 2x + 1\).

Find Critical Points for (b)

Set \(\frac{dy}{dx} = 0\), so \(x^2 - 2x + 1 = 0\). This factors to \((x-1)^2 = 0\), giving \(x = 1\).

Second Derivative Test for (b)

Find the second derivative: \(\frac{d^2y}{dx^2} = 2x - 2\). At \(x = 1\), \(\frac{d^2y}{dx^2} = 0\), indicating an inflection point, as it does not confirm a maxima or minima.

Differentiate the Function (c)

Differentiate \(y = -x^3 + 4.5x^2 - 6x + 6\) to get \(\frac{dy}{dx} = -3x^2 + 9x - 6\).

Find Critical Points for (c)

Set \(\frac{dy}{dx} = 0\): \(-3x^2 + 9x - 6 = 0\). Divide by -3: \(x^2 - 3x + 2 = 0\). This factors to \((x-1)(x-2) = 0\), giving \(x = 1\) and \(x = 2\).

Second Derivative Test for (c)

Find the second derivative: \(\frac{d^2y}{dx^2} = -6x + 9\). At \(x = 1\), \(\frac{d^2y}{dx^2} = 3\) (positive), showing a local minimum. At \(x = 2\), \(\frac{d^2y}{dx^2} = -3\) (negative), showing a local maximum.

Key concepts

Stationary Points

Stationary points of a function occur where the first derivative of the function becomes zero. In simpler terms, these are the points where the slope of the tangent to the curve is flat or horizontal. To find stationary points, you should:

• Differentiate the function to get its first derivative.
• Set this derivative equal to zero and solve for the variable of interest.

Stationary points can represent local maxima, local minima, or inflection points, depending on the behavior of the graph at those points. In the provided exercises for functions (a), (b), and (c), by setting the first derivatives to zero, we identified possible locations for these stationary points.

Second Derivative Test

After finding the stationary points using the first derivative, it's essential to determine their nature. The second derivative test helps us understand whether a stationary point is a maximum, a minimum, or an inflection point. Here's how it works:

• Compute the second derivative of the function.
• Evaluate the second derivative at each stationary point.

If the second derivative at a stationary point is:

• Positive, then it's a local minimum, as the slope changes from negative to positive.
• Negative, then it's a local maximum, indicating a slope change from positive to negative.
• Zero, the test is inconclusive, and further investigation is required to determine the point's nature.

In exercise (a), the second derivative test showed that the point at \(x=1\) was a minimum. However, in exercise (b), the second derivative was zero at \(x=1\), suggesting an inflection point. Exercise (c) revealed a mixture of outcomes at different stationary points.

Critical Points

Critical points extend beyond just stationary points. They include any points where a function's derivative is zero or undefined, meaning:

• Stationary points where the derivative equals zero, confirming a flat slope.
• Singular points where the derivative does not exist or becomes infinite.

Finding critical points involves setting the function's first derivative equal to zero and considering any points within the function's domain where the derivatives might be undefined. In these exercises, we identified critical points by solving for \(x\) values where the first derivative became zero, allowing us to pinpoint potential extrema or points of inflection in the given domain.

Differentiation

Differentiation is the process of finding the derivative of a function, which essentially measures how the function's output changes concerning changes in its input. This technique is central to calculus and allows us to:

• Determine rates of change and instantaneous velocity.
• Identify slopes of tangent lines to curves at given points.
• Find critical and stationary points by analyzing where derivatives equal zero or are undefined.

In these exercises, differentiation was used to obtain the first and second derivatives of the given functions. This enabled us to analyze the curves' behavior and locate key features like stationary points and their nature on the interval \((0, \infty)\). Understanding differentiation is fundamental in studying calculus, providing necessary tools for exploring various mathematical and real-world phenomena.