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Chapter 3: Problem 20

Find the slope of the curve at the point indicated. \(y=x^{3}-2 x+7, \quad x=-2\)

Short Answer

Expert verified
The slope of the curve at \( x = -2 \) is 10.

Step by step solution

01

Differentiate the Function

To find the slope of the curve at a given point, we need to find the derivative of the function.The function is given as \( y = x^3 - 2x + 7 \).The derivative of \( y \), denoted as \( y' \) or \( \frac{dy}{dx} \), with respect to \( x \), is:\[ \frac{dy}{dx} = 3x^2 - 2 \].
02

Evaluate the Derivative at the Given Point

Now that we have the derivative, we need to evaluate it at \( x = -2 \) to find the slope of the curve at this point.Substitute \( x = -2 \) into the derivative:\[ \frac{dy}{dx}(-2) = 3(-2)^2 - 2 \].Calculate:\[ 3(4) - 2 = 12 - 2 = 10 \].
03

Interpret the Result

The value \( 10 \) we calculated in Step 2 is the slope of the curve at the point where \( x = -2 \). This means that at this point, the curve rises 10 units for every unit it moves to the right.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Derivative
A derivative represents the rate at which a function changes at any given point. In simple terms, it tells us how steep a curve is at a specific location. The mathematical symbolism for a derivative is often written as \( f'(x) \) or \( \frac{dy}{dx} \). Derivatives are central to calculus as they allow you to compute slopes of
  • linear equations, which are constant slopes, and
  • non-linear equations, where the slope might vary from one point to another.
A derivative helps pinpoint these variable slopes, making it crucial for understanding changes in any kind of rate-dependent scenario, like speed or growth over time.
Slope of a Curve
Visualizing the slope of a curve can be compared to the concept of a tangent touching a curve at a single point. Imagine standing on a hill; the gradient of the hill at your location is akin to the slope of a curve at that point. The slope tells us
  • whether the curve is increasing or decreasing, and
  • how quickly, quantified by the numerical value of the slope.
For instance, a slope of 10, as found at \( x = -2 \) in our exercise, indicates that the curve increases steeply at this point.
Differentiation
Differentiation is the mathematical process of finding a derivative. When you differentiate a function, you essentially determine how the function changes with respect to its variable. For polynomials, this often involves applying the power rule, where:
  • The derivative of \( x^n \) is \( nx^{n-1} \).
Using differentiation, you can convert a function into its derivative form, transforming, for instance, \( y = x^3 - 2x + 7 \) into \( \frac{dy}{dx} = 3x^2 - 2 \). This new expression provides valuable information about the function's rate of change.
Evaluate Derivative
Evaluating a derivative involves substituting a specific value into the derivative function to calculate the slope at that point. This step bridges the abstract derivative concept to practical application. For instance, by substituting \( x = -2 \) into \( \frac{dy}{dx} = 3x^2 - 2 \), you compute
  • \( 3(-2)^2 - 2 = 12 - 2 = 10 \).
This result shows the curve's slope at \( x = -2 \), offering tangible insight into how quickly the function changes right then and there.
Polynomial Function
Polynomial functions are expressions composed of variables raised to whole-number exponents. These are some of the simplest types of functions to differentiate. Consider the equation \( y = x^3 - 2x + 7 \).
  • The degree of this polynomial is 3, determined by the highest exponent, which is often reflective of the curve's complexity.
  • Polynomial derivatives frequently diminish in degree by one, simplifying the function while revealing critical details about its slope at various points.
Understanding polynomial functions is foundational for grasping broader calculus concepts, as they offer a straightforward context in which to practice finding and applying derivatives.

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Most popular questions from this chapter

Find the derivatives of the functions in Exercises \(19-40\) $$ y=(5-2 x)^{-3}+\frac{1}{8}\left(\frac{2}{x}+1\right)^{4} $$

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