Question

Use a graphing calculator or a CAS Draw the graph of \(y=f(x)=x^{3}-2 x^{2}+1 .\) Then find the slope of the tangent line at (a) \(-1\) (b) 0 (c) \(\overline{1}\) (d) \(3.2\)

Short answer

Slopes: 7, 0, -1, 17.92 at x = -1, 0, 1, 3.2 respectively.

Step-by-step solution

Understanding the Function

The function given is \( f(x) = x^3 - 2x^2 + 1 \). We need to find the slope of the tangent line to this curve at various points \( x = -1, 0, 1, \) and \( 3.2 \). The slope of the tangent line at any point is given by the derivative of the function evaluated at that point.

Calculate the Derivative

Find the derivative of the function \( f(x) = x^3 - 2x^2 + 1 \). Using standard differentiation rules:\( f'(x) = \frac{d}{dx}(x^3) - \frac{d}{dx}(2x^2) + \frac{d}{dx}(1) = 3x^2 - 4x \).

Evaluate the Derivative at x = -1

Substitute \( x = -1 \) into the derivative \( f'(x) = 3x^2 - 4x \):\( f'(-1) = 3(-1)^2 - 4(-1) = 3 + 4 = 7 \).

Evaluate the Derivative at x = 0

Substitute \( x = 0 \) into the derivative \( f'(x) = 3x^2 - 4x \):\( f'(0) = 3(0)^2 - 4(0) = 0 \).

Evaluate the Derivative at x = 1

Substitute \( x = 1 \) into the derivative \( f'(x) = 3x^2 - 4x \):\( f'(1) = 3(1)^2 - 4(1) = 3 - 4 = -1 \).

Evaluate the Derivative at x = 3.2

Substitute \( x = 3.2 \) into the derivative \( f'(x) = 3x^2 - 4x \):\( f'(3.2) = 3(3.2)^2 - 4(3.2) = 3(10.24) - 12.8 = 30.72 - 12.8 = 17.92 \).

Key concepts

tangent line

When we talk about a tangent line, we are discussing a straight line that just touches a curve at a specific point. The importance of this line is that it provides a snapshot of the behavior of the curve at that location. Think of it as a tiny microcosm of the curve—so tiny that from where the tangent line begins, it perfectly matches the slope of the curve.
The beauty of tangent lines is their ability to simplify complex curves into straightforward, linear relationships just at that borne point.

• They provide the "instantaneous rate of change" of the function.
• Used heavily in physics to compute things like velocity and engineering for stress analysis.
• Depicted as a straight line just barely grazing the curve, almost like a gentle kiss.

Understanding tangent lines is crucial for calculus students, as it bridges algebraic concepts with more advanced calculus principles.

slope of a curve

The slope of a curve at any given point is a measure of how steep or gentle the tangent line is at that point. You could think of it as the tilt of the curve, and it can provide valuable insights into the curve's behavior such as when it is increasing or decreasing.
In numerically expressing this slope, we're referring to the derivative of the function. For our exercise with function \( f(x) = x^3 - 2x^2 + 1 \), the derivative \( f'(x) = 3x^2 - 4x \) captures precisely how the curve's slope changes as we glance across different \( x \)-values.

• Positive slope: The curve is going upwards.
• Negative slope: The curve is headed downwards.
• Zero slope: The curve is flat—typically where maxima, minima, or inflection points occur.

By evaluating the derivative at particular points, we can find the exact, instantaneous slope at those spots on the curve.

graphing calculator

A graphing calculator is an invaluable tool that allows students and professionals to visualize mathematical functions quickly and easily. It can draw the graph of a function like \( f(x) = x^3 - 2x^2 + 1 \), and help find key characteristics like intercepts and perhaps most notably, the behavior of tangent lines.
With just a few clicks, you can:

• Plot the curve, and visually ascertain where it peaks or dips.
• Identify points of interest like intersections with axes.
• Superimpose tangent lines to observe their slopes visually at different points.

Graphing calculators demystify complex calculus problems by transforming abstract equations into visible, comprehensible data, making calculus more approachable and less intimidating.

calculus problems

Calculus problems often involve finding derivatives, studying rates, and understanding slopes—the very essence of examining change. They can range from finding the slope of a curve to optimizing functions based on their peaks and troughs.
In this exercise, we're tasked with finding slopes at various points. This involves differentiating the given function to find \( f'(x) \), which is the key to unlocking the tangent line slopes.

• Sensible and step-by-step approach uncovers the solutions methodically.
• Numerous real-world applications, from travel to data prediction.
• Encourages logical thinking and problem-solving skills.

Through tackling calculus problems, students not only develop calculus skills but also a deeper appreciation for how change and rates influence many aspects of the world around us.