Problem 59 - Applied Mathematics: For the Managerial, Life, and Social Sciences

Question

Find the function \(f\) given that the slope of the tangent line to the graph of \(f\) at any point \((x, f(x))\) is \(f^{\prime}(x)\) and that the graph of \(f\) passes through the given point. $$f^{\prime}(x)=\frac{1}{2} x^{-1 / 2} ;(2, \sqrt{2})$$ 60\. $$f^{\prime}(t)=t^{2}-2 t+3 ;(1,2)$$

Step-by-Step Solution

Answer

The functions corresponding to the given conditions are: 1. \(f(x) = x^{\frac{1}{2}}+ (\sqrt{2} - 2^{\frac{1}{2}})\) 2. \(f(t) = \frac{1}{3}t^3 - t^2 + 3t + (2 - (\frac{1}{3} - 1 + 3))\)

Step 1: Derivative, antiderivative and the given point

Given: \(f^{\prime}(x) = \frac{1}{2}x^{-\frac{1}{2}}\) and the point \((2,√{2})\). #Step 1: Integrate the derivative# We need to find the antiderivative for \(f^{\prime}(x)\): $$ f(x) = \int f^{\prime}(x) dx = \int \frac{1}{2}x^{-\frac{1}{2}} dx $$ #Step 2: Finding the antiderivative# Using the power rule of integration, we get: $$ f(x) = (x^{\frac{1}{2} + 1}\cdot \frac{1}{\frac{1}{2} + 1} \cdot \frac{1}{2}) + C = x^{\frac{1}{2}} + C $$ where C is a constant of integration. #Step 3: Use the given point to find C# Substitute the point \((2, \sqrt{2})\) into the antiderivative and solve for C: $$ \sqrt{2} = (2)^{\frac{1}{2}} + C $$ Now, we find the value of C: $$ C = \sqrt{2} - 2^{\frac{1}{2}} $$ #Step 4: Write the final function# The original function is: $$ f(x) = x^{\frac{1}{2}}+ (\sqrt{2} - 2^{\frac{1}{2}}) $$ #Exercise 2:#

Step 2: Derivative, antiderivative and the given point

Given: \(f^{\prime}(t) = t^2 - 2t + 3\) and the point \((1, 2)\). #Step 1: Integrate the derivative# We need to find the antiderivative for \(f^{\prime}(t)\): $$ f(t) = \int f^{\prime}(t) dt = \int (t^2 - 2t + 3) dt $$ #Step 2: Finding the antiderivative# Using the power rule of integration, we get: $$ f(t) = (\frac{1}{3}t^3 - t^2 + 3t) + C $$ where C is a constant of integration. #Step 3: Use the given point to find C# Substitute the point \((1, 2)\) into the antiderivative and solve for C: $$ 2 = (\frac{1}{3}(1)^3 - (1)^2 + 3(1)) + C $$ Now, we find the value of C: $$ C = 2 - (\frac{1}{3} - 1 + 3) $$ #Step 4: Write the final function# The original function is: $$ f(t) = \frac{1}{3}t^3 - t^2 + 3t + (2 - (\frac{1}{3} - 1 + 3)) $$

Key Concepts

Integration

In calculus, integration is like the reverse process of differentiation. If differentiation tells us the rate of change, integration helps us find the function from this rate. Think of it like piecing a puzzle back together. When you're given a derivative, you're looking at how the function behaves. Integration is how we discover the original function that behaves in that way.
Understanding integration is crucial as it plays a significant role in finding areas under curves, solving differential equations, and more. It's about moving from knowing how fast something is changing to understanding what it actually is.

Antiderivative

The antiderivative, also known as an indefinite integral, is a function whose derivative is the original function. Finding an antiderivative is like figuring out the 'parent' function of the derivative.
For instance, if you have a derivative like \( f'(x) \), the antiderivative \( f(x) \) could be a function where \( f'(x) \) describes its slope at any point.

• The antiderivative's power is that it gives us a family of functions all differing by a constant, \( C \).
• This constant \( C \) comes from the fact that there could be different starting points, or initial conditions, leading to the same rate of change.

Function finding

Function finding involves determining the exact function from its derivative by integrating it. This includes finding any constants involved using known values, or 'given points.'
By substituting these known points into the integrated function, we can solve for these constants. This step translates the family of possible functions into the specific one that fits those values.

• Your goal is to find \( f(x) \) or \( f(t) \) that not only has the right rate of change but also passes through a specified point.
• This process uses both algebra and integration knowledge to solve for exact solutions.

Power rule of integration

The power rule of integration is a simple yet essential tool for finding antiderivatives. If you remember the power rule for differentiation, this is its counterpart for integration.
When you have a term like \( x^n \), the power rule states that its integral is \( \frac{x^{n+1}}{n+1} + C \).

• This works for any real number \( n \) except \( n = -1 \) since that is a special case (resulting in a logarithmic function).
• To employ the rule, just increase the power by one and divide by the new power, then don't forget to add the constant \( C \).

Mastering this rule simplifies the process of integrating polynomial functions and is fundamental to tackling more complex integrals efficiently.