Question

(a) For f(x)=2x^3, find an equation of the linear function that best fits f at x=1.

(b) Use the tangent line equation you found in (a) to approximate f(1.1).

(c) Find the actual value of f(1.1) by using the function f(x).

(d) Fill in the blank with < or >. Tangent line approx. ________ Actual value. What does this tell you about the concavity of f(x)? Explain.

Answer 1

a) f'(x) = d(2x^3)/dx = 6x^2

This is the slope of the tangent line at any point.

For x = 1, slope = 6(1)^2 = 6
At this point, x = 1, y = 2

So, using the equation of a line,

2 = m(1) + c
=> 2 = 6(1) + c
=> c = -4

So,

y - 2 = m(x - 1) + c
=> y - 2 = 6(x - 1) - 4

Answer 2

b)

f(x + Δx) = f(x) + (f'(x))*Δx

Here, x = 1, Δx = 0.1, f(x) = 2

=> f(1.1) = 2 + (6)(0.1)
= 2 + 0.6
= 2.6

c) f(1.1) = 2*(1.1)^3
= 2*(1.33)
= 2.66

d) Using the above solution, which should it be?

Answer 3

having c = -4, that gives us
y = 6x-4
using the point slope form, we get
y-2 = 6(x-1)
y = 6x-4