Problem 5 - Applied Calculus: For Business, Economics, and the Social and Life Sciences

Question

In Exercises 1 through 10 , determine intervals of increase and decrease and intervals of concavity for the given function. Then sketch the graph of the function. Be sure to show all key features such as intercepts, asymptotes, high and low points, points of inflection, cusps, and vertical tangents. $$ f(t)=3 t^{5}-20 t^{3} $$

Step-by-Step Solution

Answer

The function increases on \((-\infty, -2) \ \cup \ (2, \infty)\), decreases on \((-2, 0) \ \cup \ (0, 2)\), is concave up on \((-\infty, -\sqrt{2}) \ \cup \ (0, \sqrt{2})\ \cup \ \(2, \ \infty) \), and concave down on \((-\sqrt{2}, 0) \ \cup \ (\sqrt{2}, 2)\).

Step 1 - Find the first derivative

Determine the first derivative of the function to identify the intervals of increase and decrease. Given the function \( f(t) = 3t^5 - 20t^3 \), the first derivative is: \[ f'(t) = \frac{d}{dt}(3t^5 - 20t^3) = 15t^4 - 60t^2 \]

Step 2 - Find critical points

To find critical points, set the first derivative f'(t) equal to zero and solve for t: \[ 15t^4 - 60t^2 = 0 \] Factor out common terms: \[ 15t^2(t^2 - 4) = 0 \] Solve for t: \[ t^2 = 0 \ or \ t^2 = 4 \ which gives t = 0, t = 2, and t = -2 \]

Step 3 - Determine intervals of increase and decrease

Test the intervals between critical points \((-∞, -2), (-2, 0), (0, 2), (2, ∞)\) in f'(t): \[ f'(-3) = 15(-3)^4 - 60(-3)^2 = 540 > 0 \] increasing \[ f'(-1) = 15(-1)^4 - 60(-1)^2 = -45 < 0 \] decreasing \[ f'(1) = 15(1)^4 - 60(1)^2 = -45 < 0 \] decreasing \[ f'(3) = 15(3)^4 - 60(3)^2 = 540 > 0 \] increasing

Step 4 - Find the second derivative

Determine the second derivative to identify intervals of concavity. First derivative: \[ f'(t) = 15t^4 - 60t^2 \] Second derivative: \[ f''(t) = \frac{d}{dt}(15t^4 - 60t^2) = 60t^3 - 120t \] Factor out common terms: \[ 60t(t^2 - 2) = 0 \]

Step 5 - Find inflection points

Set the second derivative f''(t) equal to zero and solve for t: \[ 60t(t^2 - 2) = 0 \] Solve for t: \[ t = 0, t = √2, t = -√2 \]

Step 6 - Determine intervals of concavity

Test the intervals between inflection points \((-∞, -√2), (-√2, 0), (0, √2), (√2, ∞)\) in f''(t): \[ f''(-2) = 60(-2)^3 - 120(-2) = -480 < 0 \] concave down \[ f''(-1) = 60(-1)^3 - 120(-1) = 60 > 0 \] concave up \[ f''(1) = 60(1)^3 - 120(1) = -60 < 0 \] concave down \[ f''(2) = 60(2)^3 - 120(2) = 480 > 0 \] concave up

Step 7 - Identify key features

High and low points are the critical points. Check f(t) at t = 0, t = 2, t = -2: \[ f(0) = 0 \] \[ f(2) = 3(2)^5 - 20(2)^3 = 32 \] \[ f(-2) = 3(-2)^5 - 20(-2)^3 = -32 \] No vertical asymptotes or cusps since domain is all real numbers. Points of inflection at t = 0, t = √2, t = -√2.

Step 8 - Sketch the graph

Plot the points found: \(0,0\), \(2,32\), \(-2,-32\), and inflection points \(√2, f(√2)\), \(-√2, f(-√2)\). Note increasing and decreasing intervals, and concave up/down regions. This will show the overall shape and key features of the graph.

Key Concepts

derivative

The derivative of a function helps us understand how the function is changing at any given point. For the function \(f(t) = 3t^5 - 20t^3\), we first find the derivative, denoted as \(f'(t)\). This involves differentiating each term of the function separately. The result is the first derivative: \(f'(t) = 15t^4 - 60t^2\). This derivative allows us to analyze the function's behavior, such as where it is increasing or decreasing.

critical points

Critical points occur where the first derivative of the function equals zero or is undefined. For our function, we set \(f'(t) = 15t^4 - 60t^2\) equal to zero and solve for \(t\). Factoring out common terms, we get \(15t^2(t^2 - 4) = 0\), which simplifies to \(t = 0, t = 2, t = -2\). These critical points are crucial as they indicate potential maximums, minimums, or points of inflection.

concavity

Concavity tells us whether the graph of our function is curving upwards or downwards. We determine this by finding the second derivative of the function. For \(f(t) = 3t^5 - 20t^3\), the second derivative is \(f''(t) = 60t^3 - 120t\). If \(f''(t) > 0\), the function is concave up. If \(f''(t) < 0\), the function is concave down. Factoring out common terms, we get \(60t(t^2 - 2) = 0\), giving inflection points at \(t = 0, t = \sqrt{2}, t = -\sqrt{2}\).

inflection points

Inflection points occur where the concavity of the function changes. We find these points by setting the second derivative \(f''(t) = 60t^3 - 120t\) equal to zero. Solving for \(t\), we get \(t = 0, \sqrt{2}, -\sqrt{2}\). Testing intervals around these points helps us determine where the function changes concavity. For example, the function is concave down in the interval \((-∞, -\sqrt{2})\) and concave up in the interval \((-\sqrt{2}, 0)\).

graph sketching

Sketching the graph involves combining all the information found from derivatives, critical points, and inflection points. We plot key points such as where the function increases or decreases and where it changes concavity. Intervals of increase are \((-∞, -2)\) and \((2, ∞)\), while decrease intervals are \((-2, 0)\) and \((0, 2)\). Points of inflection and key features are plotted, such as \((0,0)\), \((2,32)\), and \((-2,-32)\). The final graph exhibits the overall behavior and key features of the function.