Problem 3 - Calculus with Concepts in Calculus

Question

Find the intervals on which the graph of the function is concave upward and those on which it is concave downward. $$ f(x)=x^{3}-6 x^{2}+12 x-4 $$

Step-by-Step Solution

Answer

The function is concave downward on \((-\infty, 2)\) and concave upward on \((2, \infty)\).

Step 1: Find the First Derivative

To determine where the function changes concavity, we first need to find the first derivative of the function \( f(x) \). Given \( f(x) = x^{3} - 6x^{2} + 12x - 4 \), we use the power rule to differentiate each term: \( f'(x) = 3x^{2} - 12x + 12 \).

Step 2: Find the Second Derivative

The second derivative tells us about the concavity of the function. We differentiate \( f'(x) = 3x^{2} - 12x + 12 \) to find \( f''(x) \). By applying the power rule again, we get \( f''(x) = 6x - 12 \).

Step 3: Find Critical Points of the Second Derivative

Set the second derivative equal to zero to find critical points. Solve \( 6x - 12 = 0 \) for \( x \). This gives \( 6x = 12 \) or \( x = 2 \). These are potential inflection points.

Step 4: Determine Intervals for Concavity

We test on intervals around our critical point \( x = 2 \). For \( x < 2 \), choose \( x = 1 \): \( f''(1) = 6(1) - 12 = -6 \), which is negative (concave down). For \( x > 2 \), choose \( x = 3 \): \( f''(3) = 6(3) - 12 = 6 \), which is positive (concave up).

Step 5: Conclude Intervals of Concavity

From our tests in Step 4, the function is concave downward on \((-\infty, 2)\) and concave upward on \((2, \infty)\). Thus, the intervals for concavity and concavity change at \( x = 2 \).

Key Concepts

Understanding the First Derivative

The first derivative of a function, denoted as \( f'(x) \), provides crucial information about the slope of the tangent line to the curve at any given point. When you calculate the first derivative, you're essentially determining the rate of change of the function. This helps us to easily identify where the function is increasing or decreasing. In the given exercise, we started with the function \( f(x) = x^{3} - 6x^{2} + 12x - 4 \) and found the first derivative to be \( f'(x) = 3x^{2} - 12x + 12 \). This was achieved using the power rule, where you multiply each term by its power and decrease the power by one.

• Use the first derivative to find critical points of the function by setting \( f'(x) = 0 \).
• Critical points help in analyzing and understanding function behavior.

By knowing the first derivative, we lay the groundwork for determining concavity and locating inflection points.

Diving into the Second Derivative

The second derivative, often symbolized as \( f''(x) \), gives us insight into the concavity of a function. If you imagine a curve, concavity tells you whether it opens upwards or downwards. In simple terms:

• If \( f''(x) > 0 \), the function is concave upward or shaped like a 'U'.
• If \( f''(x) < 0 \), the function is concave downward or shaped like an 'n'.

To find the concavity, calculate the derivative of the first derivative. For our function \( f(x) \), after differentiating \( f'(x) = 3x^{2} - 12x + 12 \), we obtained \( f''(x) = 6x - 12 \). This step uses similar techniques as before, differentiating once more.The role of the second derivative extends beyond just determining concavity; it also aids in better understanding points where the curve changes concavity, called inflection points. These changes are critical in the graph's shape analysis.

Exploring Inflection Points

Inflection points are fascinating, as they mark where a function changes from concave upward to concave downward or vice versa. These points provide insight into the overall structure and behavior of the curve.To find inflection points:

• Set the second derivative \( f''(x) \) equal to zero and solve for \( x \). In our example, \( 6x - 12 = 0 \) gives us \( x = 2 \). This is a potential inflection point.
• Check values on either side of \( x = 2 \) to confirm changes in concavity.

Test intervals before and after this point: - For \( x < 2 \), plugging any value into \( f''(x) \) confirms negative concavity (concave down).- For \( x > 2 \), the value is positive (concave up).Thus, the inflection point at \( x = 2 \) is where the function transitions from concave down to concave up, altering the graph's shape. Recognizing these points aids in sketching and interpreting graphs more accurately.