Exam 1 solutions

Of course! Ask away—whether it's about linear algebra, calculus, or anything math-related, I'm happy to help.

To find the minimum value of $f(x) = x^3 - 4x - 1$ over the interval $[0,3]$, we follow these steps:

Step 1: Find the derivative

The critical points occur where $f'(x) = 0$, so we compute:

Setting this equal to zero:

Since we are only considering $x$ in the interval $[0,3]$, we take the positive root:

Step 2: Evaluate $f(x)$ at critical points and endpoints

We now compute $f(x)$ at $x = 0$, $x = 3$, and $x = \frac{2}{\sqrt{3}}$.

1. At $x = 0$:

   $$f(0) = 0^3 - 4(0) - 1 = -1$$

2. At $x = 3$:

   $$f(3) = 3^3 - 4(3) - 1 = 27 - 12 - 1 = 14$$

3. At $x = \frac{2}{\sqrt{3}}$:

   $$f\left(\frac{2}{\sqrt{3}}\right) = \left(\frac{2}{\sqrt{3}}\right)^3 - 4\left(\frac{2}{\sqrt{3}}\right) - 1$$

   We compute this value numerically:

   $$\left(\frac{2}{\sqrt{3}}\right)^3 = \frac{8}{3\sqrt{3}} \approx 0.154$$ $$4 \times \frac{2}{\sqrt{3}} = \frac{8}{\sqrt{3}} \approx 4.618$$ $$f\left(\frac{2}{\sqrt{3}}\right) \approx 0.154 - 4.618 - 1 = -5.464$$

Step 3: Identify the minimum value

The function values at the critical points and endpoints are:

• $f(0) = -1$
• $f(3) = 14$
• $f( \frac{2}{\sqrt{3}} ) \approx -5.464$

Since $-5.464$ is the smallest value, the minimum value of $f(x)$ on $[0,3]$ is approximately $-5.464$, occurring at $x = \frac{2}{\sqrt{3}}$.

To find the least squares regression line $f(x) = ax + b$ for the data points $(0,0)$, $(1,0)$, and $(2,2)$, we follow these steps:

Step 1: Define the Total Squared Error Function

The least squares error function measures the sum of the squared differences between the predicted values $f(x_i)$ and the actual data values $y_i$. It is given by:

For our data points $(0,0), (1,0), (2,2)$, this becomes:

Step 2: Compute Partial Derivatives

Derivative with respect to $a$:

Derivative with respect to $b$:

Step 3: Solve for $a$ and $b$

Setting both derivatives equal to zero:

1. $10a + 6b - 8 = 0$
2. $6a + 6b - 4 = 0$

Solve the system

From equation (2):

Solving for $b$:

Substituting into equation (1):

Substituting $a = 1$ into $b = \frac{2}{3} - a$:

Step 4: Interpret the Solution

The least squares regression line is:

This line best approximates the given data points in the least squares sense.

Here's the graph of the least squares regression line $f(x) = x - \frac{1}{3}$, along with the given data points $(0,0)$, $(1,0)$, and $(2,2)$. Let me know if you need any modifications! ​​

We will use elementary row operations to transform the given matrix into reduced row echelon form (RREF). The matrix is:

Step 1: Make the first pivot 1

We want the first pivot to be 1. Since the leading coefficient in the first row is $-2$, we divide the first row by $-2$:

Step 2: Make the first column below the pivot zero

We use row operations to eliminate the first column entries below the pivot:

• $R_2 \gets R_2 - 3R_1$
• $R_3 \gets R_3 - 7R_1$

Step 3: Make the second pivot 1

We divide the second row by $\frac{1}{2}$ to make the pivot 1:

Step 4: Make the second column below and above the pivot zero

We eliminate the nonzero entries in the second column using:

• $R_1 \gets R_1 - \frac{1}{2}R_2$
• $R_3 \gets R_3 + \frac{1}{2}R_2$

Step 5: Make the third pivot 1

We divide the third row by 4:

Step 6: Make the third column above the pivot zero

We eliminate the nonzero entries above the third pivot:

• $R_1 \gets R_1 + R_3$
• $R_2 \gets R_2 - 2R_3$

Final Answer

The matrix is now in reduced row echelon form (RREF):

This is the identity matrix, which means that the original matrix is row equivalent to the identity matrix and is invertible.

You're right to double-check! The correct reduced row echelon form (RREF) of the matrix is:

This means that I made an error in my row operations. The third row should have been all zeros rather than producing the identity matrix. The correct interpretation is that the system has a dependent variable, meaning one of the columns does not contribute to a unique solution.

Would you like me to go through the corrected row reduction step-by-step? ​​

Let's carefully go through the row reduction again, ensuring each step is correct.

We start with the given matrix:

Step 1: Make the first pivot 1

We want to make the first pivot (top-left entry) a 1. Since the first entry is $-2$, we divide the first row by $-2$:

Step 2: Make the first column below the pivot zero

We eliminate the entries below the pivot using:

• $R_2 \gets R_2 - 3R_1$
• $R_3 \gets R_3 - 7R_1$

Step 3: Make the second pivot 1

We divide the second row by $\frac{1}{2}$ to make the pivot 1:

Step 4: Make the second column below and above the pivot zero

We eliminate the nonzero entries in the second column using:

• $R_1 \gets R_1 - \frac{1}{2}R_2$
• $R_3 \gets R_3 + \frac{1}{2}R_2$

Final Answer

The correct reduced row echelon form (RREF) is:

This result means that the third column is dependent on the first two columns, indicating that the system does not have a unique solution for all three variables. Instead, we can express one variable in terms of the others.

Let me know if you have any questions or if you'd like a further breakdown of any step!

Great idea! Let's go through the row reduction again while avoiding fractions by using more strategic row operations.

We start with the given matrix:

Step 1: Swap rows if needed

The first pivot should ideally be 1, but we have $-2$ in the first row. Instead of dividing, we swap $R_1$ and $R_2$ to bring a 3 to the top:

Now, we will use row operations to eliminate the first column below the pivot.

Step 2: Eliminate the first column below the pivot

We use:

• $R_2 \gets 3R_2 + 2R_1$
• $R_3 \gets 3R_3 - 7R_1$

Step 3: Make the second pivot 1

The second pivot is currently $-1$, so we multiply $R_2$ by $-1$ to make it 1:

Now, we eliminate the third row's second column entry using:

Step 4: Make the first row’s second column zero

We eliminate the first row’s second column entry using:

Step 5: Make the first pivot 1

Finally, we divide $R_1$ by 3 to get a leading 1:

Final Answer

The correct reduced row echelon form (RREF) is:

This approach successfully avoids fractions until the final step! Let me know if you want any further clarifications.

Awesome! Send over the next problem, and I'll be happy to help.

We need to find the solutions to $A\mathbf{x} = \mathbf{0}$ for the given reduced row echelon form (RREF) of $A$:

Let $\mathbf{x} = \begin{bmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \end{bmatrix}$. The system of equations corresponding to this matrix is:

Step 1: Express Leading Variables in Terms of Free Variables

From the third equation, we get:

From the first equation:

From the second equation:

Since $x_3$ is not a leading variable, we set it as a free parameter, say $t$:

Using this, we express the other variables in terms of $t$:

Step 2: Write the Solution in Vector Form

where $t$ is a free parameter in $\mathbb{R}$.

Final Answer:

The solution set for $A\mathbf{x} = \mathbf{0}$ is:

This means the null space of $A$ is spanned by the single vector $\begin{bmatrix} -1 \\ -3 \\ 1 \\ 0 \end{bmatrix}$, indicating that the null space is one-dimensional.

Let me know if anything needs further clarification! 😊

I see that you've uploaded an image. How can I assist you with it? Are you looking for an analysis, interpretation, or something specific related to the image?

Let's analyze the contour plot step by step.

Part (a): Identify Maxima, Minima, and Saddle Points

1. Minima:

   • The darkest region in the contour plot represents the lowest function values (since darker shades indicate lower values).
   • There is a clear minimum around $(-1,1)$, as this area is enclosed by concentric dark contours, indicating a local minimum.

2. Maxima:

   • The lightest regions represent the highest function values.
   • There are two local maxima, one around $(2,2)$ and another around $(1,-0.5)$, where the contours form enclosed light regions.

3. Saddle Points:

   • Saddle points occur where the function transitions between increasing and decreasing behavior in different directions.
   • There is likely a saddle point near $(0, 0.5)$, where the contours indicate a mix of increasing and decreasing behavior in different directions.

Part (b): Draw Gradient Vectors at Given Points

The gradient vector $\nabla f(x,y)$ always points in the direction of the steepest ascent and is perpendicular to contour lines.

1. Direction of Gradient Vectors:

   • If a point is in a region where the function increases toward a maximum, the gradient vector should point toward the lighter region.
   • If a point is in a region where the function decreases toward a minimum, the gradient vector should point toward the darker region.
   • At a saddle point, the gradient will be zero or point toward a nearby increase or decrease.

2. Relative Magnitude:

   • The steeper the contour lines near a point, the larger the gradient vector.
   • If the contour lines are far apart, the gradient is smaller.

Next Steps

I can draw the gradient vectors for the points marked in your contour plot to visualize this. Let me generate that for you.

Here is the contour plot with the gradient vectors (red arrows) added at the marked points:

• Each gradient vector points perpendicular to the contour lines and in the direction of increasing function values.
• The steepness of the gradient is reflected by the relative length of the arrows (longer arrows in regions where the function changes rapidly).
• Notice that at the saddle point near $(0, 0)$, the gradient is pointing vertically, as expected in such a region.

Let me know if you'd like any modifications or further explanations! 😊 ​​

Here is the contour plot with the identified critical points:

• Red (X): Local maxima at approximately $(2,2)$ and $(1,-0.5)$.
• Blue (X): Local minimum at approximately $(-1,1)$.
• Green (X): Saddle point near $(0, 0.5)$.

This visualization should now accurately reflect the structure of the function represented by the contour plot. Let me know if any adjustments are needed! 😊 ​​

I can't directly remove images from our conversation, but you can restart the chat or manually clear your conversation history before sharing. Let me know if you need me to summarize anything again without images! 😊