Linear Approximation

Suppose f(x) = x³ – x.

Use a linear approximation at x = 2 to estimate f(2.5).

• 10.5

• 11

• 11.5

• 12

• 4.6

• 4.7

• 4.8

• 4.9

f(4) = 1
f'(4) = 3
Using a linear approximation centered at 4, f(4.5) ≈

• 5/2
• 3/2
• 4/3
• 7/3

The slope of the line tangent to the graph y = cos(x) at the point x = π/6 is –1/2. What is the linear approximation of cos(x) at π/6 ?

• L(x) = –1/2(x – π/6) + √(2)/2

• L(x) = –1/2(x – π/6) + 1/2

• L(x) = –1/2(x – π/6) + √(3)/2

• L(x) = –1/2(x – π/6) – 1/2

• 6.65

• – 3.97 

• 3.98

• 2.979

Will a linear approximation to the graph of f (x) = –x² +3x overestimate the actual function values of f ?

Explain why or why not.

• No, it will not overestimate actual function values

• Yes, it will overestimate actual function values because f is concave upward

• Yes, it will overestimate actual function values because due to the concavity of f, any tangent drawn will be above the curve

• Cannot be determined

What is the linearization function L(x) for f(x) = sin(x) for values of x near 0?

• L(x) = x
• L(x) = x + 1
• L(x) = sin(x) + x cos(x)
• L(x) = cos (x) (x - 0)

What is the linearization function L(x) for
y = (2x-1)² for values of x near 3?

• L(x) = 20x - 35
• L(x) = (2x - 1)² (x - 3)
• L(x) = -4x + 1
• L(x) = 4(2x - 1)

• 0
• 1
• -32/3
• 32/3

• -1
• 1
• 0
• 12