Problem 76 - 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. $$ f(x)=6 x^{5}-45 x^{4}+80 x^{3}+30 x^{2}-30 x+2 $$

Step-by-Step Solution

Answer

Critical points and inflection points can be found using numerical methods; behavior is determined via second derivative test.

Step 1: Find the First Derivative

To find the critical points, we need to compute the first derivative of the function. Take the derivative of each term: \[ f'(x) = \frac{d}{dx}(6x^5 - 45x^4 + 80x^3 + 30x^2 - 30x + 2) \]Calculating the derivative results in:\[ f'(x) = 30x^4 - 180x^3 + 240x^2 + 60x - 30 \]

Step 2: Solve for Critical Points

Critical points occur where the first derivative is equal to zero. Solve the equation:\[ 30x^4 - 180x^3 + 240x^2 + 60x - 30 = 0 \]Use a numerical solver or graphing tool to approximate the solutions. These solutions give the critical points.

Step 3: Analyze Behavior at Critical Points

Calculate the second derivative to determine the behavior at each critical point:\[ f''(x) = \frac{d}{dx}(30x^4 - 180x^3 + 240x^2 + 60x - 30) \]This results in:\[ f''(x) = 120x^3 - 540x^2 + 480x + 60 \]Substitute the critical points into the second derivative to determine whether each is a local minimum, maximum, or saddle point.

Step 4: Find the Inflection Points

Inflection points occur where the second derivative changes sign. Set the second derivative equal to zero:\[ 120x^3 - 540x^2 + 480x + 60 = 0 \]Solve this equation using a numerical solver or graphing tool to approximate the inflection points. Verify each point by checking a sign change.

Key Concepts

Critical Points

In the realm of calculus, critical points are key values of a function where its rate of change switches behavior. These points usually occur where the first derivative, denoted as \( f'(x) \), equals zero or is undefined. For the function given, \[ f(x)=6 x^{5}-45 x^{4}+80 x^{3}+30 x^{2}-30 x+2, \] we calculate the first derivative, and from there, solve for when that derivative equals zero:

• This means solving \( f'(x) = 30x^4 - 180x^3 + 240x^2 + 60x - 30 = 0 \).

Finding these roots might require numerical methods or technology because analytical solutions can be cumbersome. Each solution is a critical point. These points help us predict the slope's direction change and behavior at certain points of the curve.

Inflection Points

Inflection points are locations where a curve changes concavity, either from concave up to down or vice versa. Essentially, these are the points where the second derivative — \( f''(x) \) — changes its sign.

• These are determined by solving \( f''(x) = 120x^3 - 540x^2 + 480x + 60 = 0 \).

At an inflection point, the second derivative will cross the x-axis. The value of \( x \) where this happens indicates a shift in the curvature direction. Solving for these values can require numerical methods or graphing tools. Confirming a sign change of the second derivative across these points ensures they are true points of inflection. Recognizing where inflection points occur is crucial for understanding the overall shape and behavior of the graph.

First Derivative

The first derivative of a function provides us with the slope or gradient of the function at any given point. For our function, the derivative is:\[ f'(x) = 30x^4 - 180x^3 + 240x^2 + 60x - 30 \] The power of the first derivative lies in its ability to indicate where functions increase or decrease:

• If \( f'(x) > 0 \), the function is increasing at that interval.
• If \( f'(x) < 0 \), the function is decreasing at that interval.

Moreover, by finding the roots of \( f'(x) \), we pinpoint critical points, where potential local maxima, minima, or saddle points occur. This insight is key to mapping the fluctuations in the function's graph.

Second Derivative

The second derivative is a powerful tool that provides information about the acceleration of a curve, which directly relates to the concavity of the graph. For our function, the second derivative is derived as follows:\[ f''(x) = 120x^3 - 540x^2 + 480x + 60 \] The sign of \( f''(x) \) tells us the concavity:

• If \( f''(x) > 0 \), the function is concave up (like a smile) in that region, indicating a local minimum if this is a critical point.
• If \( f''(x) < 0 \), the function is concave down (like a frown), pointing to a local maximum at any critical points.

Totaling the foundational information from both the first and second derivatives provides a comprehensive view of a function's graph, showing turning points, as well as shifts in concavity.