(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.

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

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?

having c = -4, that gives us

y = 6x-4
using the point slope form, we get
y-2 = 6(x-1)
y = 6x-4

(a) To find an equation of the linear function that best fits f at x=1, we can use the concept of a tangent line. The slope of the tangent line at a given point on the curve of a function represents the instantaneous rate of change of the function at that point. The equation of a straight line in slope-intercept form is y = mx + b, where m represents the slope and b represents the y-intercept.

In this case, we need to find the slope of the tangent line to f(x)=2x^3 at x=1. To do this, we can take the derivative of f(x) with respect to x. The derivative of f(x) = 2x^3 is f'(x) = 6x^2.

Evaluating the derivative at x=1, we find f'(1) = 6(1)^2 = 6. This gives us the slope of the tangent line at x=1.

Now we can use the point-slope form of a line to find the equation of the tangent line. The equation is given by y - y1 = m(x - x1), where (x1, y1) represents the point (1, f(1)) and m is the slope.

Since we are looking for the equation at x=1, we have (x1, y1) = (1, f(1)), but f(1) is not directly provided. So let's find f(1) first.

(b) To approximate f(1.1) using the tangent line equation found in (a), we substitute x=1.1 into the equation of the tangent line.

(c) To find the actual value of f(1.1) using the function f(x) = 2x^3, we substitute x=1.1 into the function.

(d) Comparing the tangent line approximation and the actual value of f(1.1), we need to determine if the tangent line approximation (<) is less than or greater than the actual value. This comparison gives us insights into the concavity of f(x).

To understand the concavity of f(x), we can examine the difference between the tangent line approximation and the actual value. If the tangent line approximation is less than the actual value, it suggests that the function is concave up. Conversely, if the tangent line approximation is greater than the actual value, it suggests that the function is concave down.

By comparing the tangent line approximation and the actual value, we can determine whether the function f(x) = 2x^3 is concave up or concave down at x = 1.1.