Problem 61 - Calculus: Early Transcendental Functions

Question

Find a function \(f(x)\) such that the point (1,2) is on the graph of \(y=f(x),\) the slope of the tangent line at (1,2) is 3 and \(f^{\prime \prime}(x)=x-1\)

Step-by-Step Solution

Answer

The solution is \(f(x) = \frac{1}{6}x^3-0.5x^2+3.5x-1.17\).

Step 1: Integrate to Find the Derivative

Integrate the given function \(f^{\prime \prime}(x)=x-1\) to find the derivative \(f^{\prime}(x)\). This gives \(f^{\prime}(x)= 0.5x^2-x+C_1\), where \(C_1\) is the constant of integration.

Step 2: Use Given Condition to Find \(C_1\)

Use the fact that the slope of the tangent line at (1,2) is 3, which means \(f^{\prime}(1) = 3\). Substituting into our expression for \(f^{\prime}(x)\), we get \(3 = 0.5*1^2-1+C_1\), which solves to \(C_1=3.5\). So \(f^{\prime}(x) = 0.5x^2-x+3.5\).

Step 3: Integrate to Find the Function

Integrate the derivative to find the function \(f(x)\): \(f(x)= \frac{1}{6}x^3-0.5x^2+3.5x+C_2\), where \(C_2\) is the new constant of integration.

Step 4: Use The Given Condition to Find \(C_2\)

Use the fact that the point (1,2) is on the graph, which means \(f(1)=2\). Substituting into our function, we get \(2=\frac{1}{6}*1^3-0.5*1^2+3.5*1+C_2\), which solves to \(C_2=-1.17\). So, our function is \(f(x)=\frac{1}{6}x^3-0.5x^2+3.5x-1.17\).

Key Concepts

Understanding Tangent Lines

In calculus, a tangent line is a line that touches a curve at a single point without crossing it. This concept is crucial for understanding how functions behave locally. To find the equation of a tangent line at a specific point, you need to know both the point of tangency and the slope of the tangent line at that point. The slope of the tangent line is equivalent to the derivative of the function at the given point. For example, in the original exercise, the slope of the tangent line at the point (1,2) is given to be 3. This information allows us to adjust our derivative equation to ensure that the slope matches with the given condition. By substituting this point into the derivative, we can solve for any unknown constants.

Exploring Second Derivatives

The second derivative represents the rate of change of the first derivative, or in simpler terms, the curvature of the graph. It gives us insight into the "concavity" of a function.

• If the second derivative is positive, the graph is concave up, resembling a "U" shape.
• If it's negative, the graph is concave down, resembling an upside-down "U".

In our exercise, the second derivative is given as \(f^{\prime \prime}(x) = x-1\). This tells us how the slope of the tangent lines changes. For example, if you were to plug in values of \(x\), you could determine how quickly or slowly the slope increases or decreases at various points. Integrating the second derivative gives us the first derivative, which we can then further refine using initial conditions.

The Role of Constants of Integration

When finding antiderivatives, constants of integration often appear. These constants reflect the fact that there is an infinite family of functions that could have produced the given derivative.

• The first constant of integration, \(C_1\), is determined by conditions related to the first derivative, like the slope of a tangent line at a specific point.
• The second constant of integration, \(C_2\), is determined by conditions related to the actual function, like a specific point lying on the function's graph.

In the provided exercise, you find \(C_1\) to ensure that the slope at the point (1,2) matches the given condition \(f^\prime(1) = 3\). Similarly, \(C_2\) is found by ensuring the function passes through the point (1,2), i.e., \(f(1) = 2\). These constants are vital because they tailor the general solution of an integral to the specific problem at hand.