Problem 16 - Precalculus Mathematics for Calculus

Question

Find the derivative of the function at the given number. $$f(x)=2-3 x+x^{2} \quad \text { at }-1$$

Step-by-Step Solution

Answer

The derivative at \( x = -1 \) is \( -5 \).

Step 1: Write Down the Function

Let's start by considering the function given, which is a quadratic function: \[ f(x) = 2 - 3x + x^2 \] We need to find its derivative.

Step 2: Differentiate the Function

We need to differentiate \( f(x) = 2 - 3x + x^2 \) with respect to \( x \). Applying the power rule (\( n \cdot x^{n-1} \)), we differentiate each term:- The derivative of the constant \( 2 \) is \( 0 \).- The derivative of \( -3x \) is \( -3 \cdot x^{1-1} = -3 \).- The derivative of \( x^2 \) is \( 2x^{2-1} = 2x \).Thus, the derivative \( f'(x) \) is:\[ f'(x) = 0 - 3 + 2x = 2x - 3 \]

Step 3: Evaluate the Derivative at the Given Number

Now, we need to evaluate the derivative \( f'(x) = 2x - 3 \) at \( x = -1 \). Substitute \( x = -1 \) into the derivative:\[ f'(-1) = 2(-1) - 3 \]Calculate \( f'(-1) \):\[ f'(-1) = -2 - 3 = -5 \]

Step 4: Final Step: Write Down the Solution

The derivative of the function \( f(x) = 2 - 3x + x^2 \) at \( x = -1 \) is \( -5 \).

Key Concepts

Quadratic Function

When dealing with a quadratic function, you're encountering one of the simplest types of polynomial functions. A quadratic function has the general form \( ax^2 + bx + c \), where \( a \), \( b \), and \( c \) are constants and \( a eq 0 \). These functions create a parabola when graphed on a coordinate plane.

The basic features of a parabola include its vertex, axis of symmetry, and direction of opening. The vertex is the highest or lowest point on the graph, depending on whether the function opens upwards (\( a > 0 \)) or downwards (\( a < 0 \)). For the given function \( f(x) = 2 - 3x + x^2 \), it's a standard quadratic function. As you can see, this function creates a parabola that opens upwards since the coefficient of \( x^2 \) is positive.

Understanding these basic properties provides a solid foundation for further calculus concepts, such as differentiation.

Power Rule

The power rule is a fundamental concept in calculus, used for finding derivatives of polynomial functions easily. It states that the derivative of \( x^n \) is \( n \cdot x^{n-1} \). This rule simplifies the process by allowing you to systematically work through polynomial terms.

In our given example, \( f(x) = 2 - 3x + x^2 \), applying the power rule:

• The derivative of the constant \( 2 \) becomes \( 0 \) because constants disappear when differentiated.
• The derivative of \( -3x \) is \( -3 \), as you multiply by the exponent \( 1 \) and reduce it by one (\( x^{1-1} \)).
• The derivative of \( x^2 \) is \( 2x \), multiplying by the exponent \( 2 \) and reducing it to \( x^{2-1} \).

So, the power rule guides us to find the derivative \( f'(x) = 2x - 3 \). This step is crucial in understanding how individual function components contribute to the behavior of the entire function.

Evaluate Derivative

Evaluating the derivative at a specific point allows us to understand the function's behavior in a particular spot. It's like zooming in on the function to see how it changes at that instant.

For the function \( f(x) = 2 - 3x + x^2 \), after differentiating, you get \( f'(x) = 2x - 3 \).
You are tasked with evaluating this derivative at \( x = -1 \).

To do this, plug \( -1 \) into \( f'(x) \):

• First, substitute \( -1 \) into the derivative: \( f'(-1) = 2(-1) - 3 \).
• Next, calculate the result: \(-2 - 3 = -5 \).

This calculation shows that the rate of change, or slope, of the function at \( x = -1 \) is \( -5 \). This means that at \( x = -1 \), the function is decreasing steeply. Evaluating derivatives helps in identifying not just slopes but also finding critical points and analyzing the behavior of a function in different contexts.