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Ch. 3 - Derivatives
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077Not the one you use?Change textbook
All textbooksHass 15th EditionCh. 3 - DerivativesProblem 3.75
Chapter 3, Problem 3.75

Find the points on the curve y = 2x³ - 3x² - 12x + 20 where the tangent line is


a. perpendicular to the line y = 1 - (x/24).
_
b. parallel to the line y = √2 - 12x.

Verified step by step guidance
1
To find the points where the tangent line is perpendicular to the line y = 1 - (x/24), first determine the slope of this line. The slope is -1/24.
For perpendicularity, the slope of the tangent line to the curve y = 2x³ - 3x² - 12x + 20 must be the negative reciprocal of -1/24, which is 24.
Find the derivative of the curve y = 2x³ - 3x² - 12x + 20 to get the slope of the tangent line: \( \frac{dy}{dx} = 6x^2 - 6x - 12 \). Set this equal to 24 and solve for x.
To find the points where the tangent line is parallel to the line y = √2 - 12x, determine the slope of this line, which is -12.
Set the derivative \( \frac{dy}{dx} = 6x^2 - 6x - 12 \) equal to -12 and solve for x to find the points where the tangent line is parallel.

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Key Concepts

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

Derivative

The derivative of a function measures the rate at which the function's value changes at a given point. It is the slope of the tangent line to the curve at that point. To find points where the tangent line is perpendicular or parallel to another line, we need to calculate the derivative of the curve and set it equal to the appropriate slope derived from the given lines.
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Derivatives

Slope of a Line

The slope of a line is a measure of its steepness, calculated as the change in y over the change in x (rise over run). For a line in the form y = mx + b, 'm' represents the slope. When determining where the tangent line to the curve is parallel or perpendicular to another line, we compare the slopes: parallel lines have equal slopes, while perpendicular lines have slopes that are negative reciprocals of each other.
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Critical Points

Critical points occur where the derivative of a function is zero or undefined, indicating potential locations for local maxima, minima, or points of inflection. In this context, finding critical points helps identify where the tangent line to the curve has specific slopes, which is essential for solving the problem of finding points where the tangent is parallel or perpendicular to the given lines.
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Related Practice
Textbook Question

Differential Estimates of Change


In Exercises 35–40, write a differential formula that estimates the given change in volume or surface area.


The change in the lateral surface area S = πr√(r² + h²) of a right circular cone when the radius changes from r₀ to r₀ + dr and the height does not change

Textbook Question

In Exercises 83–88, find equations for the lines that are tangent, and the lines that are normal, to the curve at the given point.


(y - x)² = 2x + 4, (6, 2)

Textbook Question

a. Graph the function


ƒ(x) = { x², -1 ≤ x < 0

{ -x², 0 ≤ x ≤ 1.


b. Is ƒ continuous at x = 0?

c. Is ƒ differentiable at x = 0?


Give reasons for your answers.

Textbook Question

Find the derivatives of the functions in Exercises 19–40.


s = (4 / 3π)sin(3t) + (4 / 5π)cos(5t)

Textbook Question

In Exercises 41–58, find dy/dt.


y = tan²(sin³(t))

Textbook Question

Is there a value of b that will make


g(x) = { x + b, x < 0

cos x, x ≥ 0


continuous at x = 0? Differentiable at x = 0? Give reasons for your answers.