Problem 20 - 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: Understand the Problem

We need to find the derivative of the function \( f(x) = 2 - 3x + x^2 \) and evaluate this derivative at the point \( x = -1 \).

Step 2: Differentiate the Function

To differentiate the function \( f(x) = 2 - 3x + x^2 \), we apply the power rule to each term. The derivative of a constant is 0. The derivative of \(-3x\) is \(-3\), and the derivative of \(x^2\) is \(2x\). Thus, the derivative of the function \( f(x) \) is \( f'(x) = -3 + 2x \).

Step 3: Substitute the Given Value

Now that we have the derivative \( f'(x) = -3 + 2x \), substitute \( x = -1 \) into this expression to find the derivative at the given point. That is, \( f'(-1) = -3 + 2(-1) \).

Step 4: Calculate the Result

Compute \( f'(-1) = -3 + 2(-1) = -3 - 2 = -5 \).

Key Concepts

Power Rule

The power rule is a fundamental concept when dealing with derivatives, especially when working with polynomial functions. It provides a quick way to find the derivative of terms of the form \(x^n\). The power rule states that for any real number \(n\), the derivative of \(x^n\) is given by \( \frac{d}{dx}[x^n] = nx^{n-1} \). This means you simply multiply the coefficient by the power and then decrease the power by one.

Let's apply this to the function \(f(x) = 2 - 3x + x^2\):

• The first term, "2," is a constant, so its derivative is 0.
• The second term, "-3x," involves \(x\) to the power of 1, so its derivative using the power rule is \(-3 \times 1 = -3\).
• For the \(x^2\) term, the derivative is \(2 \times x^{2-1} = 2x\).

By applying these calculations, we find that the derivative of the function \(f(x)\) is \(f'(x) = -3 + 2x\). This concise method saves time and reduces errors when solving such problems.

Differentiation

Differentiation is the process of finding the derivative of a function. The derivative provides significant information about the behavior of a function, such as its rate of change or the slope of the tangent line at any given point. In calculus, differentiation is a core technique used to analyze and understand functions deeply.

For our given function $f(x) = 2 - 3x + x^2$, we've successfully differentiated it into $f'(x) = -3 + 2x$. This new function, the derivative, tells us how the original function changes as $x$ changes. It allows us to understand the function's dynamic properties and how each of its components contributes to this change.

Another critical aspect of differentiation is recognizing that the derivative's value at a certain point offers the slope of the original function at that point, which leads us to our next step: evaluating derivatives for specific values.

Substitution in Functions

Substitution in functions involves replacing variables with specific values to find the outcome of a function under certain conditions. This concept is particularly useful in evaluating derivatives at specific points to determine things like slope or rate of change exactly where we need it.

In the previously discussed function $f(x) = 2 - 3x + x^2$, after differentiating, we obtained $f'(x) = -3 + 2x$. To find the derivative specifically at $x = -1$, we use substitution:

• Substitute -1 into $f'(x)$: $f'(-1) = -3 + 2(-1)$
• Calculate the result: $f'(-1) = -3 - 2 = -5$

This tells us that at $x = -1$, the slope of the tangent to the function is -5. Substitution not only helps solve for tangible results, but also provides insight into how the function behaves locally around specific points.