Ch. 3 - Derivatives

Briggs - Calculus: Early Transcendentals 3rd Edition

Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Not the one you use?Change textbook

Briggs 3rd Edition

Ch. 3 - Derivatives

Problem 15a

Chapter 3, Problem 15a

Use definition (1) (p. 133) to find the slope of the line tangent to the graph of f at P.
f(x) = x² - 5; P(3,4)

Verified step by step guidance

Step 1: Recall the definition of the derivative as the slope of the tangent line at a point. The derivative of a function \( f(x) \) at a point \( x = a \) is given by the limit \( f'(a) = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h} \).

Step 2: Identify the function \( f(x) = x^2 - 5 \) and the point \( P(3, 4) \). We need to find the derivative \( f'(x) \) and evaluate it at \( x = 3 \).

Step 3: Substitute \( f(x) = x^2 - 5 \) into the derivative definition: \( f'(x) = \lim_{h \to 0} \frac{(x+h)^2 - 5 - (x^2 - 5)}{h} \).

Step 4: Simplify the expression inside the limit: \( (x+h)^2 - 5 - (x^2 - 5) = x^2 + 2xh + h^2 - 5 - x^2 + 5 = 2xh + h^2 \).

Step 5: Factor out \( h \) from the numerator: \( \frac{2xh + h^2}{h} = \frac{h(2x + h)}{h} = 2x + h \). Now, take the limit as \( h \to 0 \): \( f'(x) = \lim_{h \to 0} (2x + h) = 2x \). Evaluate \( f'(3) = 2(3) = 6 \).

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Tangent Line

A tangent line to a curve at a given point is a straight line that touches the curve at that point without crossing it. The slope of the tangent line represents the instantaneous rate of change of the function at that point, which is crucial for understanding how the function behaves locally.

Derivative

The derivative of a function at a point quantifies how the function's output changes as its input changes. It is defined as the limit of the average rate of change of the function as the interval approaches zero. For the function f(x) = x² - 5, the derivative will provide the slope of the tangent line at any point on the graph.

Point of Tangency

The point of tangency is the specific point on the curve where the tangent line touches. In this case, point P(3,4) is where we need to evaluate the function and its derivative to find the slope of the tangent line. This point is essential for applying the derivative to determine the slope accurately.

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Related Practice

Textbook Question

Find f′(x) if f(x) = 15e^3x.

Textbook Question

Suppose the position of an object moving horizontally along a line after t seconds is given by the following functions s = f(t), where s is measured in feet, with s > 0 corresponding to positions right of the origin.

Determine the velocity and acceleration of the object at t = 1.

f(t) = t² − 4t; 0 ≤ t ≤ 5

Textbook Question

On what intervals is the speed increasing?

f(t) = t² - 4t; 0 ≤ t ≤ 5

Textbook Question

5–24. For each of the following composite functions, find an inner function u=g(x) and an outer function y=f(u) such that y=f(g(x)). Then calculate dy/dx.

y = (3x+7)¹⁰

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Textbook Question

7–14. Find the derivative the following ways:

a. Using the Product Rule (Exercises 7–10) or the Quotient Rule (Exercises 11–14). Simplify your result.

y = x² - 2ax +a² / x-a, where a is a constant

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Textbook Question

Determine the acceleration of the object when its velocity is zero.

f(t) = t² - 4t; 0 ≤ t ≤ 5