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

Question

(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