Problem 15 - Calculus Early Transcendentals

Question

Find all critical points of the following functions. $$f(x, y)=x^{4}-2 x^{2}+y^{2}-4 y+5$$

Step-by-Step Solution

Answer

Question: Find the critical points of the function $$f(x, y) = x^{4} - 2 x^{2} + y^{2} - 4 y + 5$$. Answer: The critical points of the function are $(0, 2)$, $(1, 2)$, and $(-1, 2)$.

Step 1: Calculate the partial derivatives of the function

To begin, let's find the partial derivatives of $$f(x, y)$$ with respect to x and y: $$\frac{\partial f(x, y)}{\partial x} = 4x^{3} - 4x$$ $$\frac{\partial f(x, y)}{\partial y} = 2y - 4$$

Step 2: Solve for x and y when partial derivatives are zero

Now we need to solve these equations to find the critical points: $$4x^{3} - 4x = 0$$ $$2y - 4 = 0$$ From the second equation, we find: $$y = 2$$ Let's focus on the first equation, we can factor out a $$4x$$ term: $$4x (x^{2} - 1)=0$$ This equation has three possible solutions: $$x = 0$$ $$x = 1$$ $$x = -1$$

Step 3: Write down the critical points

We found three values for x and one value for y. Therefore, there are three critical points for the given function: $$(x, y) = (0, 2)$$ $$(x, y) = (1, 2)$$ $$(x, y) = (-1, 2)$$ These are the critical points of the function $$f(x, y) = x^{4} - 2 x^{2} + y^{2} - 4 y + 5$$.

Key Concepts

Partial Derivatives

Partial derivatives are essential tools in multivariable calculus. They are used to measure how a function changes as each variable changes, keeping the other variables constant. For any function of two variables, let's denote it as \( f(x, y) \), partial derivatives provide insights into its surface.
For the function \( f(x, y) = x^{4} - 2x^{2} + y^{2} - 4y + 5 \), we compute the partial derivatives with respect to \( x \) and \( y \).- **Partial derivative with respect to x**: - Differentiate all terms of the function considering \( y \) as a constant. - This yields \( \frac{\partial f}{\partial x} = 4x^3 - 4x \).
- **Partial derivative with respect to y**: - Here, treat \( x \) as a constant, and differentiate all terms involving \( y \). - This results in \( \frac{\partial f}{\partial y} = 2y - 4 \).Partial derivatives are foundational in finding critical points, where these derivatives equal zero, suggesting a point of equilibrium.

Solve Equations

After determining the partial derivatives, we move ahead to find where these expressions are equal to zero. This helps us locate the critical points of a function. It is essential to solve each partial derivative equation independently.- For the partial derivative \( \frac{\partial f}{\partial x} = 4x^{3} - 4x \), we set it to zero: - Factor this equation: \( 4x(x^2 - 1) = 0 \). - This gives \( x = 0 \), \( x = 1 \), and \( x = -1 \).
- Similarly, solve \( \frac{\partial f}{\partial y} = 2y - 4 = 0 \). - Simple algebra leads to \( y = 2 \).The solution for \( y \) remains constant while \( x \) provides multiple values. Solving these equations individually finds all points where both spatial dimensions flatten simultaneously, critical to identifying places where extreme values can occur.

Function Analysis

Function analysis involves understanding the behavior of a function through its structure and derived elements like derivatives. By locating critical points, we gain valuable insights into the function's nature and how it behaves under different inputs.- **Critical Points**: - Defined as points where the gradient \( abla f = (\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}) \) is zero. - It indicates potential maximums, minimums, or saddle points.
- For the function \( f(x, y) = x^{4} - 2x^{2} + y^{2} - 4y + 5 \): - The critical points determined are \((0, 2) \), \((1, 2) \), and \((-1, 2)\).Understanding these suggests shapes or contours on the graph of the function, helping in sketching a rough surface map. By noting where these critical points lie, students or analysts can anticipate the broader implications on practical, real-world scenarios involving changes in multiple dimensions.