Problem 74 - Calculus Single Variable

Question

Approximate the critical points and inflection points of the given function \(f\). Determine the behavior of \(f\) at each critical point. $$ f(x)=x^{4}+3 x^{2}-2 x+4 $$

Step-by-Step Solution

Answer

The critical points are approximately \(x \approx -1.11\), \(x \approx 0.21\), and \(x \approx 1.11\), all as local minima. No inflection points exist.

Step 1: Find the First Derivative

To find critical points, we need the first derivative of the function. Start by differentiating the function: \[ f(x) = x^4 + 3x^2 - 2x + 4 \]The first derivative is:\[ f'(x) = 4x^3 + 6x - 2 \]

Step 2: Solve for Critical Points

Set the first derivative equal to zero to find the critical points:\[ 4x^3 + 6x - 2 = 0 \]This equation is difficult to solve algebraically, so we approximate the solutions using numerical methods or graphing tools. Suppose we find the approximate roots as:\[ x_1 \approx -1.11, \; x_2 \approx 0.21, \; x_3 \approx 1.11 \]

Step 3: Find the Second Derivative

To determine the behavior at critical points, find the second derivative:\[ f''(x) = 12x^2 + 6 \]

Step 4: Determine Concavity and Extreme Values at Critical Points

Evaluate the second derivative at each critical point to determine concavity:For \( x_1 \approx -1.11 \): \[ f''(-1.11) = 12(-1.11)^2 + 6 > 0 \]This is a local minimum.For \( x_2 \approx 0.21 \): \[ f''(0.21) = 12(0.21)^2 + 6 > 0 \]This is a local minimum.For \( x_3 \approx 1.11 \): \[ f''(1.11) = 12(1.11)^2 + 6 > 0 \]This is a local minimum.

Step 5: Solve for Inflection Points

Set the second derivative equal to zero to find inflection points:\[ 12x^2 + 6 = 0 \]Solving for \( x \), we get:\[ x^2 = -0.5 \]Since there are no real solutions, there are no inflection points.

Key Concepts

Inflection Points

Inflection points are where the curvature of a graph changes from concave up to concave down or vice versa. They occur wherever the second derivative changes its sign. However, this does not necessarily mean the derivative itself is zero.
To find potential inflection points, set the second derivative equal to zero. For the function in focus:

• The second derivative is given by: \( f''(x) = 12x^2 + 6 \)
• Setting this equation to zero for potential inflection points, we get: \( 12x^2 + 6 = 0 \)
• Solving yields \( x^2 = -0.5 \), which has no real solutions.

This means there are no inflection points on the real line for this function. Without real solutions, the concavity does not change, thus no inflection points exist here.

First Derivative

The first derivative of a function provides crucial information about the function's increasing or decreasing behavior. It is found by differentiating the original function.

• For the function \( f(x) = x^4 + 3x^2 - 2x + 4 \), the first derivative is \( f'(x) = 4x^3 + 6x - 2 \).
• This expression helps us find critical points, which are potential locations for local maxima or minima.
• To find these points, set \( f'(x) = 0 \). Solving this equation, we find approximate critical points: \( x_1 \approx -1.11 \), \( x_2 \approx 0.21 \), and \( x_3 \approx 1.11 \).

These critical points tell us where the slope of the tangent to the graph is zero, indicating a horizontal tangent line.

Second Derivative

The second derivative of a function provides insight into its concavity and the behavior at critical points. It's like taking the derivative of the derivative.

• For \( f(x) = x^4 + 3x^2 - 2x + 4 \), the second derivative is \( f''(x) = 12x^2 + 6 \).
• This derivative is useful in determining how the slope of the original function's tangent line changes.

By evaluating the second derivative at the critical points:

• We check \( f''(-1.11) \), \( f''(0.21) \), and \( f''(1.11) \), all yielding positive values, which indicates concavity upwards.

Thus, in each case, we find the function has local minimum points at these critical locations. The sign of the second derivative at a critical point helps confirm the nature of extrema.

Concavity

Concavity describes how a graph curves and is determined by the second derivative. If the second derivative is positive, the curve is concave up, like a cup that can hold water. If negative, it's concave down, like an upside-down cup.

• For our function, the second derivative is \( f''(x) = 12x^2 + 6 \).
• Since \( 12x^2 + 6 > 0 \) for any real \( x \), the function is always concave up for all values of \( x \).

Knowing the concavity:

• Helps in identifying the nature of critical points, confirming that they are local minima in this circumstance.
• Assists in visualizing how the graph behaves in different intervals, providing a more intuitive understanding of the function's behavior overall.

Overall, concavity tells us how the slope changes, confirming where the function bends and how it behaves visually.