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

f(x) = x²(x+1)²
Which of these is the linear approximation to f centered at x = 1?

• L(x) = 12x - 1

• L(x) = 6x + 5

• L(x) = 12x - 8

• L(x) = 12x - 16

f(x) = x^-2/5
Use a linear approximation centered at x = 32 to approximate (30)^-2/5.

• 17/180

• 25/64

• 23/128

• 41/160

Which of the following represents the linear approximation to f (x) at the point (c, f(c)) ?

• L(x) = f'(c) + f(c)(x - c)

• L(x) = f(c) + f'(c)(x - c)

• L(x) = f'(c) + f(c)(x + c)

• L(x) = f(c) - f'(c)(x - c)

How can we minimize the error of an approximation in a linear approximation at the point (a, f (a))?

• By translating the point of tangency in either direction

• By restricting the values of f (x) sufficiently close to f (a)

• By iterating further away from the point of tangency

• By restricting the values of x sufficiently close to 'a'

• 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

Suppose f (x) = 3x² is graphed in the x-y plane. Will any linear approximations to f overestimate or underestimate the actual function values?

• overestimate

• underestimate