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Finding the Slope of a Tangent Line In Exercises \(5-10\) , find the slope of the tangent line to the graph of the function at the given point. $$ g(x)=x^{2}-9, \quad(2,-5) $$

Short Answer

Expert verified
The slope of the tangent line of the function \(g(x) = x^{2} - 9\) at the point (2, -5) is 4.

Step by step solution

01

Find the derivative of the function

First, find the derivative of the function \(g(x) = x^{2} - 9\). The derivative of the function can be found by applying the power rule of derivatives. The power rule states that if we have a function of the form \(f(x) = x^{n}\), its derivative is given by \(f'(x) = n \cdot x^{n-1}\). Applying this rule to our function, the derivative \(g'(x)\) will be \(2x\).
02

Calculate the slope of the tangent line

The slope of the tangent line to the graph of the function at a given point is the value of the derivative at that point. So, to find the slope of the tangent line at the point (2, -5), we need to substitute \(x = 2\) into the derivative \(g'(x)\). Doing this, \(g'(2) = 2 \cdot 2 = 4\). Therefore, the slope of the tangent line at the point (2, -5) is 4.

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Key Concepts

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

Derivative
The concept of a derivative is fundamental in calculus. A derivative represents the rate at which a function is changing at any given point. Imagine you have a curve graphed on a coordinate system. The derivative at a particular point on that curve gives you the slope of the line tangent to the curve at that point. This line is an approximation of the curve very close to the point.

When calculating the derivative of a function, you are essentially finding a new function, usually denoted as \(f'(x)\), which describes how \(f(x)\) changes as \(x\) changes. Therefore, knowing the value of the derivative of a function at a specific point provides information about the behavior of the function in that small neighborhood of the point.
Power Rule
The power rule is a quick and efficient technique used in calculus to find the derivative of functions of the form \(x^n\), where \(n\) is any real number. This rule simplifies the process significantly and is essential for solving many calculus problems.

To use the power rule, if you have a function, say \(f(x) = x^n\), where \(n\) is a constant, the derivative \(f'(x)\) is computed as \(n \cdot x^{n-1}\). This means you multiply the power by the coefficient (if any), and reduce the power by one.
  • It's as simple as bringing the exponent down as a multiplier.
  • Then subtracting one from the original exponent to get the new exponent.

By using the power rule, you can quickly find the derivative of polynomial expressions, like in the original problem with \(g(x) = x^2 - 9\). Here, the derivative \(g'(x)\) is calculated as \(2x\).
Function Graph
A function graph is a visual representation of a function's behavior, mapping each input \(x\) to a corresponding output \(y = f(x)\). It helps in understanding relationships and patterns in a mathematical expression.

For the function \(g(x) = x^2 - 9\), the graph is a parabola opening upwards, shifted downwards along the y-axis. The point (2, -5) is particularly important here as it lies on this curve, allowing us to explore the specific behavior of the function at that point by examining the slope of the tangent line.
  • The tangent line touches the graph at exactly one point, aligning with the curve's direction.
  • The slope of this line reflects how steeply the function is changing at that specific point.

Understanding the graph of a function bridges the gap between algebraic manipulation and geometric interpretation.
Math Problem Solving
Solving math problems is like putting the pieces of a puzzle together. It involves understanding the problem, identifying the right tools or techniques, and applying them methodically to arrive at the solution. This structured approach is critical when finding the slope of a tangent line using calculus.

Given a function and a point, like \(g(x) = x^2 - 9\) and (2, -5), you begin by determining the derivative. This informs you how the function is behaving generally. Next, substitute the specific x-value into the derivative to find how rapidly the function changes at that particular point.
  • Break down complex problems into manageable steps.
  • Use known rules, like the power rule, to simplify calculations.

This organized method ensures errors are minimal and solutions are robust, making math less intimidating and more like solving a well-crafted puzzle.

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Most popular questions from this chapter

Finding a Second Derivative In Exercises \(91-98\) , find the second derivative of the function. $$ f(x)=x \sin x $$

A trough is 12 feet long and 3 feet across the top (see figure). Its ends are isosceles triangles with altitudes of 3 feet. (a) Water is being pumped into the trough at 2 cubic feet per minute. How fast is the water level rising when the depth \(h\) is 1 foot? (b) The water is rising at a rate of \(\frac{3}{8}\) inch per minute when \(h=2 .\) Determine the rate at which water is being pumped into the trough.

A boat is pulled into a dock by means of a winch 12 feet above the deck of the boat (see figure). (a) The winch pulls in rope at a rate of 4 feet per second. Determine the speed of the boat when there is 13 feet of rope out. What happens to the speed of the boat as it gets closer to the dock? (b) Suppose the boat is moving at a constant rate of 4 feet per second. Determine the speed at which the winch pulls in rope when there is a total of 13 feet of rope out. What happens to the speed at which the winch pulls in rope as the boat gets closer to the dock?

A ladder 25 feet long is leaning against the wall of a house (see figure). The base of the ladder is pulled away from the wall at a rate of 2 feet per second. (a) How fast is the top of the ladder moving down the wall when its base is 7 feet, 15 feet, and 24 feet from the wall? (b) Consider the triangle formed by the side of the house, the ladder, and the ground. Find the rate at which the area of the triangle is changing when the base of the ladder is 7 feet from the wall. (c) Find the rate at which the angle between the ladder and the wall of the house is changing when the base of the ladder is 7 feet from the wall.

Using Related Rates In Exercises \(1-4,\) assume that \(x\) and \(y\) are both differentiable functions of \(t\) and find the required values of \(d y / d t\) and \(d x / d t .\) $$ \begin{array}{rlrl}{y=\sqrt{x}} & {\text { (a) } \frac{d y}{d t} \text { when } x=4} & {} & {\frac{d x}{d t}=3} \\ {} & {\text { (b) } \frac{d x}{d t} \text { when } x=25} & {} & {\frac{d y}{d t}=2}\end{array} $$

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