Problem 21 - Precalculus Mathematics for Calculus

Question

Find \(f^{\prime}(a),\) where \(a\) is in the domain of \(f .\) $$f(x)=x^{2}+2 x$$

Step-by-Step Solution

Answer

\( f^{\prime}(a) = 2a + 2 \)

Step 1: Differentiate the Function

To find the derivative of the function, apply the power rule. The function is given as \( f(x) = x^2 + 2x \). The power rule states that \( \frac{d}{dx}[x^n] = nx^{n-1} \). Apply the power rule to each term of the function.- The derivative of \( x^2 \) is \( 2x \).- The derivative of \( 2x \) is \( 2 \).Thus, \( f^{\prime}(x) = 2x + 2 \).

Step 2: Evaluate the Derivative at \( x = a \)

Replace \( x \) with \( a \) in the expression for the derivative found in Step 1. So, we substitute \( a \) into \( f^{\prime}(x) = 2x + 2 \), and we get:\( f^{\prime}(a) = 2a + 2 \).

Key Concepts

Power Rule

The power rule is a fundamental technique in calculus used for finding derivatives. It's incredibly useful for differentiating functions involving powers of variables like polynomials. The basic idea is straightforward: if you have a function of the form \( x^n \), then its derivative is \( nx^{n-1} \). Here, "n" is just the exponent of the variable in the function.
For example, in the function \( f(x) = x^2 + 2x \):

• The term \( x^2 \) becomes \( 2x^{2-1} = 2x \) when differentiated.
• For the term \( 2x \), think of it as \( 2x^1 \), so it becomes \( 2 \times 1 \times x^{1-1} = 2 \).

By applying the power rule, you systematically reduce the power of each term and scale by the original power. This method makes differentiating simple polynomial expressions quick and efficient. Always remember: decrease the exponent by one after multiplying by the original exponent!

Differentiation

Differentiation is the process of finding the derivative of a function. In calculus, derivatives measure how a function changes as its input changes. This concept is central because it helps us understand the rate of change and the slope of a curve at any point.
In our example, the function \( f(x) = x^2 + 2x \) is differentiated to find \( f'(x) \), which gives us the slope of the tangent to the curve at any given \( x \). The derivative here, using the power rule, results in:

• \( f'(x) = 2x + 2 \)

This tells us that for any tiny increment in \( x \), the function \( f(x) \) increases by \( 2x + 2 \) times that increment. Differentiation thus provides valuable insights into the behavior and trends of functions and their graphical representations.

Evaluation at a Point

Evaluation at a point involves substituting a specific value into a derivative to find the rate of change of the function at that particular input. This process is crucial because it gives us concrete numbers to describe the behavior of the function at specific points on its curve.
After differentiating \( f(x) = x^2 + 2x \), we found that the derivative is \( f'(x) = 2x + 2 \). To evaluate the derivative at a specific point \( x = a \), substitute \( a \) into the derivative:

• \( f'(a) = 2a + 2 \)

This result indicates the instantaneous rate of increase or decrease of the function at \( x = a \). Evaluation at specific points can answer practical questions, such as the exact steepness of a hill (if the function represents height) or the acceleration of a particle at a particular time.