Week 6: Preparation

Week 6: Preparation#

Reading Material#

  • Review: We recommend that you re-read Section 2.2 in Chapter 2

  • Reading: Chapter 5

  • Python: Demo 6

Key Concepts#

Long Day will cover:

  • The image set of continuous functions

  • Global and local extrema (maxima and minima)

  • Stationary points

  • Extremum investigation of multi-variable scalar functions

Short Day will cover:

  • Characterization of stationary points (maximum, minimum, or saddle)

  • The second-order test and the “Hessian-eigenvalue” test

  • Positive (semi-)definite matrices

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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