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Ch. 3 - Derivatives
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Not the one you use?Change textbook
All textbooksBriggs 3rd EditionCh. 3 - DerivativesProblem 3.1.23a
Chapter 3, Problem 3.1.23a

Use definition (2) (p. 135) to find the slope of the line tangent to the graph of f at P.
f(x) = 3x2 - 4x; P(1, -1)

Verified step by step guidance
1
Step 1: Recall the definition of the derivative as the slope of the tangent line at a point. The derivative of a function f at a point x is given by the limit: f'(x) = \(\lim\)_{h \(\to\) 0} \(\frac{f(x+h) - f(x)}{h}\).
Step 2: Substitute the given function f(x) = 3x^2 - 4x into the derivative definition. Calculate f(x+h) by replacing x with (x+h) in the function: f(x+h) = 3(x+h)^2 - 4(x+h).
Step 3: Expand the expression for f(x+h): f(x+h) = 3(x^2 + 2xh + h^2) - 4x - 4h = 3x^2 + 6xh + 3h^2 - 4x - 4h.
Step 4: Substitute f(x) and f(x+h) into the derivative formula: \(\frac{f(x+h) - f(x)}{h}\) = \(\frac{(3x^2 + 6xh + 3h^2 - 4x - 4h) - (3x^2 - 4x)}{h}\).
Step 5: Simplify the expression: \(\frac{6xh + 3h^2 - 4h}{h}\) = 6x + 3h - 4. Take the limit as h approaches 0 to find f'(x) = 6x - 4. Evaluate f'(1) to find the slope of the tangent line at P(1, -1).

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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.
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Slopes of Tangent Lines

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. In this context, finding the derivative of f(x) will provide the slope of the tangent line at point P.
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Point-Slope Form

The point-slope form of a linear equation is used to express the equation of a line given a point on the line and its slope. It is written as y - y₁ = m(x - x₁), where (x₁, y₁) is a point on the line and m is the slope. This form is useful for constructing the equation of the tangent line once the slope has been determined.
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Related Practice
Textbook Question

Use definition (2) (p. 135) to find the slope of the line tangent to the graph of f at P.

f(x) = 8 - 2x2; P(0, 8)

Textbook Question

21–30. Derivatives

a. Use limits to find the derivative function f' for the following functions f.

f(s) = 4s³+3s; a= -3, -1

Textbook Question

31–32. Velocity functions A projectile is fired vertically upward into the air, and its position (in feet) above the ground after t seconds is given by the function s(t).

a. For the following functions s(t), find the instantaneous velocity function v(t). (Recall that the velocity function v is the derivative of the position function s.)

s(t)= −16t²+100t

Textbook Question

Consider the following cost functions.

a. Find the average cost and marginal cost functions.

C(x) = 1000+0.1x, 0≤x≤5000, a=2000

Textbook Question

13-26 Implicit differentiation Carry out the following steps.

a. Use implicit differentiation to find dy/dx.

(x+y)^2/3=y; (4, 4)

Textbook Question

The following table gives the distance f(t) fallen by a smoke jumper seconds after she opens her chute. <IMAGE>

a. Use the forward difference quotient with ℎ = 0.5 to estimate the velocity of the smoke jumper at t=2 seconds.