Calculus: Graphical, Numerical, Algebraic

3rd Edition

ISBN: 9780133688399

Author: Finney, Ross L.

Publisher: Pearson/Prentice Hall

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Question

Chapter 2.4, Problem 34E

To determine

To find: The point at which the tangent to the curve f(x)=3−4x−x2 is horizontal.

Expert Solution & Answer

Answer to Problem 34E

The point at which the tangent to the curve f(x)=3−4x−x2 is horizontal is (−2,7).

Explanation of Solution

Given information:

The curve is f(x)=3−4x−x2.

Calculation:

The tangent to any point is horizontal if the slope of the curve at that point is equal to 0.

The formula for the slope of curve at x=a is equal to limh→0f(a+h)−f(a)h.

Substitute a+h for x in the function.

f(x)=3−4x−x2f(a+h)=3−4(a+h)−(a+h)2f(a+h)=3−4a−4h−(a2+h2+2ah)f(a+h)=3−4a−a2−h(4+2a)−h2

Substitute a for x in the function.

f(x)=3−4x−x2f(a)=3−4(a)−(a)2f(a)=3−4a−a2

Substitute 3−4a−a2−h(4+2a)−h2 for f(a+h) and 3−4a−a2 for f(a) in the formula for slope.

limh→0f(a+h)−f(a)h=limh→03−4a−a2−h(4+2a)−h2−(3−4a−a2)h=limh→0−h(4+2a)−h2h=limh→0h(−4−2a−h)h=limh→0(−4−2a−h)

Further simplify.

limh→0f(a+h)−f(a)h=−4−2a−0=−4−2a

Now, equate the slope to zero.

−4−2a=02a=−4a=−2

Substitute −2 for x in the function.

f(x)=3−4x−x2f(−2)=3−4(−2)−(−2)2f(−2)=3+8−4f(−2)=7

Therefore, the point at which the tangent to the curve f(x)=3−4x−x2 is horizontal is (−2,7).

Chapter 2.4, Problem 33E

Chapter 2.4, Problem 35E

Chapter 2 Solutions

Calculus: Graphical, Numerical, Algebraic

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The point at which the tangent to the curve f ( x ) = 3 − 4 x − x 2 is horizontal.