Problem 4 - Precalculus: Mathematical for Calculus

Question

Find the slope of the tangent line to the graph of \(f\) at the given point. $$f(x)=1+2 x-3 x^{2} \quad \text { at }(1,0)$$

Step-by-Step Solution

Answer

The slope of the tangent line at \((1, 0)\) is \(-4\).

Step 1: Understand the Problem

We need to find the slope of the tangent line to the function \( f(x) = 1 + 2x - 3x^2 \) at the point \((1, 0)\). This involves finding the derivative of the function and evaluating it at \( x = 1 \).

Step 2: Differentiate the Function

The derivative of the function \( f(x) = 1 + 2x - 3x^2 \) with respect to \( x \) is found using basic differentiation rules. \[ f'(x) = \frac{d}{dx}(1) + \frac{d}{dx}(2x) - \frac{d}{dx}(3x^2) \] \[ f'(x) = 0 + 2 - 6x \] Thus, the derivative is \( f'(x) = 2 - 6x \).

Step 3: Evaluate the Derivative at the Given Point

Now, substitute the \( x \)-value from the given point \((1, 0)\) into the derivative to find the slope of the tangent line.\[ f'(1) = 2 - 6(1) = 2 - 6 = -4 \]

Step 4: Conclusion

The slope of the tangent line to the graph of \( f(x) \) at the point \((1, 0)\) is \(-4\).

Key Concepts

Tangent Line

A tangent line is a straight line that just touches a curve at a certain point, and it represents the slope of the curve at that specific point. This means the slope of the tangent line is equal to the instantaneous rate of change of the function at that point.

• For any given curve, there are infinitely many tangent lines depending on the point you choose.
• In this exercise, we are interested in finding the tangent line at a specific point on the curve defined by the function.

Understanding tangent lines is crucial because they help us grasp how a curve changes.
In many real-world scenarios, such as physics and engineering, the tangent line can indicate trends and predict future behavior of a system.

Derivative

In calculus, a derivative represents the rate at which a function is changing at any given point. It's a fundamental tool to determine the slope of a tangent line.
When you differentiate a function, you're essentially finding a formula that gives you the slope of the function's curve at any point along its path.

• The derivative can be found by applying certain rules of differentiation such as the power rule, product rule, and chain rule, among others.
• For our function, the derivative is calculated as follows:
  Given function: \( f(x) = 1 + 2x - 3x^2 \) We find each part's derivative: \( f'(x) = \frac{d}{dx}(1) + \frac{d}{dx}(2x) - \frac{d}{dx}(3x^2) \) Steps produce: \( f'(x) = 0 + 2 - 6x \).

Derivatives help us understand how the slope of the tangent line changes across different points on a curve, providing insights into the behavior of the function.

Function Evaluation

Function evaluation involves substituting a specific value into a function to determine the outcome or specific result of the function at that point. In the context of finding a tangent line, it means plugging the specific point's x-value into the function's derivative to find the slope at that point.

• For our specific problem, we evaluated the derivative \( f'(x) = 2 - 6x \) at \( x = 1 \).
• This means replacing the x in the derivative with 1, yielding \( f'(1) = 2 - 6(1) \).

Ultimately, this step shows us the slope of the tangent line to the curve at our specified point, which is crucial for sketching or understanding the curve's behavior. Knowing how to perform function evaluations allows us to precisely analyze a curve's dynamics at any point we care about.