Problem 10 - Single Variable Calculus

Question

In Exercises \(5-38\), sketch the graph of the function using the curve- sketching guidelines on page \(348 .\) $$ f(t)=3 t^{4}+4 t^{3} $$

Step-by-Step Solution

Answer

The graph of \(f(t) = 3t^4 + 4t^3\) has x-intercepts at \(t = 0\) and \(t = -\frac{4}{3}\), a y-intercept at (0, 0), critical points at \(t = 0\) and \(t = -1\), and is concave down when \(t < -\frac{4}{3}\) and concave up when \(-\frac{4}{3} < t\). The graph approaches positive infinity as \(t \rightarrow -\infty\) and \(t \rightarrow \infty\).

Step 1: Find the x-intercepts and y-intercept

To find the x-intercepts, set the function equal to zero and solve for t. \(f(t) = 3t^4 + 4t^3 = 0\) Factor out t^3 from the equation: \(t^3(3t + 4) = 0\) Therefore, the x-intercepts are \(t = 0\) and \(t = -\frac{4}{3}\). To find the y-intercept, set t = 0 and solve for f(t): \(f(0) = 3(0)^4 + 4(0)^3 = 0\) The y-intercept is at (0, 0).

Step 2: Find the critical points

To find the critical points, compute the first derivative of the function and set it equal to zero. \(f'(t) = \frac{d}{dt}(3t^4 + 4t^3)= 12t^3 + 12t^2\) Factor out 12t^2 from the equation: \(12t^2( t + 1 ) = 0\) The critical points are \(t = 0\) and \(t = -1\).

Step 3: Determine the concavity

To determine the concavity of the function, we will compute the second derivative and analyze the sign of the result. \(f''(t) = \frac{d^2}{dt^2}(3t^4 + 4t^3) = 36t^2 + 24t\) Analyze the intervals where the second derivative changes sign: - For \(t < -\frac{4}{3}\): \(f''(t) < 0\), so the function is concave down. - For \(-\frac{4}{3} < t < 0\): \(f''(t) > 0\), so the function is concave up. - For \(t > 0\): \(f''(t) > 0\), so the function is concave up.

Step 4: Analyze the end behavior

As \(t \rightarrow -\infty\), \(3t^4\) dominates the function and, since it has an even exponent, the function approaches positive infinity: \(f(t) \rightarrow \infty\). As \(t \rightarrow \infty\), \(3t^4\) dominates the function as well, and since it has an even exponent, the function approaches positive infinity: \(f(t) \rightarrow \infty\). Now that we have analyzed the key information, we can sketch the graph of the function, keeping in mind the x-intercepts, y-intercept, critical points, concavity, and end behavior.

Key Concepts

Understanding X-intercepts in Curve Sketching

X-intercepts are the points where the graph of a function crosses the x-axis. These occur at values where the function is equal to zero. For the function given, \( f(t) = 3t^4 + 4t^3 \), we find the x-intercepts by solving the equation \( 3t^4 + 4t^3 = 0 \).
First, we factor the equation as \( t^3(3t + 4) = 0 \). This shows us that the x-intercepts occur at \( t = 0 \) and \( t = -\frac{4}{3} \).
Finding x-intercepts is crucial because it helps to understand the points where the function changes signs as it crosses the x-axis.

Determining the Y-intercept of a Function

The y-intercept is the point where the graph of the function crosses the y-axis. To find this intercept, substitute \( t = 0 \) into the function and solve for \( f(t) \).
For our function, the calculation is straightforward:

• \( f(0) = 3(0)^4 + 4(0)^3 = 0 \)

Thus, the y-intercept is located at the point \( (0, 0) \). Identifying the y-intercept is essential, as it gives insight into where the function begins relative to the origin.

Locating Critical Points

Critical points in curve sketching are points where the function's derivative is zero or undefined. These points often indicate local maximums or minimums or points of inflection. To find these for our function, we calculate its first derivative.Starting with \( f(t) = 3t^4 + 4t^3 \), the derivative \( f'(t) \) is given by

• \( f'(t) = 12t^3 + 12t^2 \)

Setting the derivative equal to zero:

• \( 12t^2(t + 1) = 0 \)

From this, we find the critical points at \( t = 0 \) and \( t = -1 \). Critical points are significant in determining where a function's growth rate changes, offering insights into its local behavior.

Performing a Concavity Analysis

Concavity analysis involves assessing the second derivative of a function to determine where the curve is concave up (cupped upwards) or concave down (cupped downwards). This helps visualize how the graph "bends" at different sections.For \( f(t) = 3t^4 + 4t^3 \), compute the second derivative:

• \( f''(t) = 36t^2 + 24t \)

Examining the sign of \( f''(t) \) reveals concavity:

• For \( t < -\frac{4}{3} \), \( f''(t) < 0 \); concave down.
• For \( -\frac{4}{3} < t < 0 \), \( f''(t) > 0 \); concave up.
• For \( t > 0 \), \( f''(t) > 0 \); concave up.

Analyzing concavity is vital in predicting and understanding the shape of the graph.

Interpreting End Behavior

End behavior in curve sketching describes how a function behaves as the input variable approaches positive or negative infinity. It indicates the direction in which the function's graph extends at the far reaches of the x-axis.For the function \( f(t) = 3t^4 + 4t^3 \), as \( t \to -\infty \) or \( t \to \infty \):
The term \( 3t^4 \) dominates because it has the highest power. With an even degree, \( 3t^4 \) ensures that:

• As \( t \to -\infty \), \( f(t) \to \infty \).
• As \( t \to \infty \), \( f(t) \to \infty \).

Understanding end behavior is crucial for grasping how the function performs at extreme values, providing a full picture of the graph's overall trajectory.