Problem 13 - Single Variable Calculus

Question

Determine where the graph of the function is concave upward and where it is concave downward. Also, find all inflection points of the function. $$ f(t)=t^{4}-2 t^{3} $$

Step-by-Step Solution

Answer

The graph of the function \(f(t) = t^4 - 2t^3\) is concave upward when \(0 < t < 1\) and \(t > 1\), concave downward when \(t < 0\), and has inflection points at \((0, 0)\) and \((1, -1)\).

Step 1: Find the first derivative of the function.

To find where the function is concave upward or downward, we need to determine the second derivative of the function. First, let's find the first derivative of the function, denoted as \(f'(t)\). We will use the power rule of differentiation: \(\frac{d}{dx}x^n = nx^{n-1}\) $$ f'(t) = \frac{d}{dt}(t^4 - 2t^3) = 4t^3 - 6t^2 $$

Step 2: Find the second derivative of the function.

Now that we've found the first derivative, let's find the second derivative, denoted as \(f''(t)\). Again, we will use the power rule of differentiation. $$ f''(t) = \frac{d}{dt}(4t^3 - 6t^2) = 12t^2 - 12t = 12t(t - 1) $$

Step 3: Find the critical points of the second derivative.

To find the inflection points and intervals of concavity, we need to find the critical points of the second derivative, which are the points where \(f''(t) = 0\) or \(f''(t)\) is undefined. In this case, \(f''(t)\) is never undefined, so we only need to solve for when \(f''(t) = 0\). $$ 12t(t - 1) = 0 $$ This equation has two solutions: \(t = 0\) and \(t = 1\). These are the critical points of the second derivative.

Step 4: Determine the intervals of concavity and inflection points.

Now that we have the critical points of the second derivative, we can use them to determine the intervals of concavity and the inflection points. For the intervals of concavity, we will test the second derivative using values in the intervals created by the critical points. 1. Test the interval \((-∞, 0)\): Choose a value, say -1, and plug it into \(f''(t)\): \(f''(-1) = -12 < 0\), so this interval is concave downward. 2. Test the interval \((0, 1)\): Choose a value, say 0.5, and plug it into \(f''(t)\): \(f''(0.5) = 3 > 0\), so this interval is concave upward. 3. Test the interval \((1, ∞)\): Choose a value, say 2, and plug it into \(f''(t)\): \(f''(2) = 24 > 0\), so this interval is concave upward. Now we can summarize our findings: Concave upward: \(0 < t < 1\) and \(t > 1\) Concave downward: \(t < 0\) For inflection points, they occur where the concavity changes. From our analysis above, we can see that the concavity changes at the critical points \(t = 0\) and \(t = 1\). Therefore, the inflection points are at \(t = 0\) and \(t = 1\). To find the corresponding function values, we plug these points back into the original function: Inflection point 1: \((0, f(0)) = (0, 0)\) Inflection point 2: \((1, f(1)) = (1, -1)\) So, we have found that the graph of the function is concave upward when \(0 < t < 1\) and \(t > 1\). It is concave downward when \(t < 0\). The inflection points are at \((0, 0)\) and \((1, -1)\).

Key Concepts

Second Derivative

The second derivative of a function gives us valuable information about the function's concavity. Mathematically, the second derivative is the differentiation of the first derivative. In simple terms, while the first derivative tells us the rate of change of a function, the second derivative provides insights into how this rate of change itself changes over time or in relation to the variable. For instance, when we calculated the second derivative of the function \[ f''(t) = 12t(t - 1) \]we were able to interpret its behavior concerning concavity. If the second derivative is positive over an interval, the function is concave upward on that interval. Conversely, if it is negative, the function is concave downward. Thus, solving for when the second derivative equals zero helps identify transitions in concavity.

Critical Points

Critical points in the context of concavity concern those points where the second derivative equals zero or does not exist, although in many well-defined mathematical functions, the latter is rare. These are crucial because they potentially mark inflection points, where a function's concavity changes. In our exercise, determining\[ f''(t) = 0 \]provided us with critical points at \(t = 0\) and \(t = 1\). By evaluating these points, you can segment the function on its domain and assess concavity within those intervals. Testing values within these subdivided intervals then lets you effectively chart where the function switches between being concave up and concave down.

Inflection Points

An inflection point shows where there is a concavity change in the graph of a function. This happens precisely at the critical points of the second derivative where the sign changes from positive to negative or vice versa. In simpler terms, the graph can shift from curving upward (concave up) to downward (concave down), or vice versa, at these points. In our example, we identified inflection points by first solving the equation\[ f''(t) = 12t(t - 1) = 0 \]This equation provided the inflection points at \(t = 0\) and \(t = 1\). We then determined the function values at these points to place them accurately on the graph, confirming concavity transitions.

Differentiation Power Rule

The differentiation power rule is a fundamental derivative rule used to find the derivative of functions expressed as powers of a variable. It's a straightforward yet powerful concept essential for calculus. According to this rule, if \(y = x^n\), then the derivative \(dy/dx = nx^{n-1}\). This principle was applied multiple times in the problem to find both the first and second derivatives efficiently. Applying this rule:- We found the first derivative: \[ f'(t) = 4t^3 - 6t^2 \]- Similarly, the second derivative was derived as: \[ f''(t) = 12t^2 - 12t \]This method simplifies complicated calculations since it breaks down differentiating powers into a more manageable formula, allowing us to quickly move towards analyzing the function's behavior.