Week 6: Preparation

Week 6: Preparation#

Reading Material#

We recommend that you read the textbook. Watching YouTube videos on the week’s topics can be useful, but it should not replace proper preparation for the week’s program and is not recommended as a standalone approach.

Read and study the following:

Key Concepts#

After reading, you should be able to explain the following key concepts:

  • Image set of continuous function

  • Extremum (minimum or maksimum)

  • Global extremum

  • Local extremum

  • Stationary points and other conditions

  • Extremum determination

  • Second-order test and Hessian matrix

  • Positive (semi-)definite matrices

This week, we will explore these key concepts in great detail. We expect you to have familiarized yourself with these topics before lectures.

Preparatory Exercises#

I: Maximum and Minimum of Functions#

Read the proposition in the first theorem of this week’s chapter.

Question a#

Find the minimum and maximum values as well as the image set of the function \(f : [-2,1] \to \mathbb{R}\) given by:

\[\begin{equation*} f(x) = x^2. \end{equation*}\]

Answer

Minimum value: \(m = 0\). Maximum value: \(M = 4\).

Answer

As the function is continuous, then according to the theorem the image set is an interval given by \(\operatorname{im}(f) = [0,4]\).

Question b#

We now consider some functions that do not fulfill the assumptions in the theorem:

\[\begin{align*} g_1 : \mathbb{R} \to \mathbb{R}, \quad g_1(x) &= x^3 \\ g_2 : \,\,]-\pi/2,\pi/2[\,\, \to \mathbb{R}, \quad g_2(x) &= \tan(x) \\ g_3 : [-1,1] \to \mathbb{R}, \quad g_3(x) &= \begin{cases} x+1 & ,\,x \in [-1,0] \\ 3/2-x & ,\,x \in \,]0,1] \end{cases} \end{align*}\]

Determine the minimum and maximum values (if they exist) as well as the image set of each of the functions \(g_i\,,\,i=1,2,3\).

II: Second-Order Test at Stationary Points#

Consider a function \(f: \mathbb{R} \to \mathbb{R}\) given by

\[\begin{equation*} f(x) = x^3 - 3x^2 + 4. \end{equation*}\]

Question a#

Find the stationary points by following this procedure:

  • Calculate the first derivative \(f'(x)\).

  • Let \(f'(x) = 0\), and solve for \(x\) in order to find the stationary points.

Hint

There are two stationary points. They can be found as the roots of a second-degree polynomial.

Answer

The first derivative:

\[\begin{equation*} f'(x) = \frac{d}{dx}(x^3 - 3x^2 + 4) = 3x^2 - 6x. \end{equation*}\]

This can be factorized to:

\[\begin{equation*} f'(x) = 3x(x - 2). \end{equation*}\]

This is set equal to zero:

\[\begin{equation*} 3x(x - 2) = 0, \end{equation*}\]

and the solution is found to be \(x = 0\) and \(x = 2\). Hence, the stationary points are found at \(x = 0\) and \(x = 2\).

Question b#

Classify each stationary point by following this procedure:

  • Calculate the second derivative \(f''(x)\).

  • Evaluate \(f''(x)\) at each stationary point:

    • If \(f''(x) > 0\), the point is a local minimum.

    • If \(f''(x) < 0\), the point is a local maximum.

    • If \(f''(x) = 0\), the test is inconclusive.

Answer

The second derivative is:

\[\begin{equation*} f''(x) = \frac{d}{dx}(3x^2 - 6x) = 6x - 6. \end{equation*}\]

Evaluating \(f''(x)\) at \(x = 0\) gives:

\[\begin{equation*} f''(0) = 6(0) - 6 = -6. \end{equation*}\]

Since \(-6 < 0\), then the point \(x = 0\) is a local maximum.

Evaluating \(f''(x)\) at \(x = 2\) gives:

\[\begin{equation*} f''(2) = 6(2) - 6 = 6. \end{equation*}\]

Since \(6 > 0\), then the point \(x = 2\) is a local minimum.

Question c#

Consider the theoretical propositions in the previous question. Sketch the graph of \(f(x)\) in Python to visually confirm your results.

Question d#

Is the local minimum you found also a global minimum? Is the local maximum you found also a global maximum?

Answer

No and no. The image set is all of \(\mathbb{R}\) since \(f\) is a third-degree polynomial, so there is no global minimum or maximum.

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