Worksheet 2-1

Worksheet 2-1#

Download: CMSE382-WS2_1.pdf

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Worksheet 2-1: Q1#

The function \(f(x)=3x^4 - 28x^3 + 84x^2 - 96x\) over \([-1,6]\) is plotted at desmos.com/calculator/nevlfz0yxp. The derivative of \(f\) is \(f'(x) = 12x^3 - 84x^2 +168x-96= 12(x-1)(x-2)(x-4)\).

  • Using the derivative, find all critical points of \(f(x)\).

  • Identify all stationary points, and for each stationary point, identify its type (min, max, saddle).

  • Find the global minimum and maximum of \(f(x)\) over the interval \([-1,6]\).

Worksheet 2-1: Q2#

Find the stationary points of \(f(x,y)=6x^2 y - 3x^3 + 2y^3 - 150y\).

(You can visually check your answer at desmos.com/3d/xa4komuwmb.)

Worksheet 2-1: Q3#

On a quiz, Dr. Munch asks about a function \(f:U \to \mathbb{R}\) defined on \(U \subseteq \mathbb{R}^n\) where all partial derivatives of \(f\) exist.

  • Kylo Ren writes: “At a local optimum, the gradient is zero, so \(\nabla f(x^*)=0\).”

    Mark his answer correct or explain why his answer is wrong.

  • Rey Skywalker writes: “Since \(\nabla f(\mathbf{x}^*)=0\), \(f\) has a local optimum.”

    Mark her answer correct or explain why her answer is wrong.

  • Poe Dameron writes: “Since \(f\) has a local optimum at \(\mathbf{x}^*\) and \(\mathbf{x}^*\) is in the interior of \(U\), the gradient is zero, meaning \(\nabla f(\mathbf{x}^*)=0\).”

    Mark his answer correct or explain why his answer is wrong.

Worksheet 2-1: Q4#

Let \(f(x,y)=2x+3y:S\to \mathbb{R}\) and \(S=B[0,1]=\{(x,y): x^2+y^2 \leq 1\}\).

(a) We call \(f\) a linear map if there is a matrix \(A\) such that \(f(x,y)=A \mathbf{x}\), where \(\mathbf{x}=\begin{bmatrix} x\\ y \end{bmatrix}\). Find the matrix \(A\) to show that \(f\) is a linear map.

(b) Note that because \(A\) is just a vector, \(A\mathbf{x}\) is the same as the dot product. Use the Cauchy-Schwarz inequality to find \(\argmin_{x \in S} f(x)\) and \(\argmax_{x \in S} f(x)\).

(c) What do the points you found in part (b) represent?

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