Let f be the function given by f(x)=2^x^2.

Question

Selected values of f are given in the table below. If the values in the table are used to approximate f′(0.5), what is the difference between the approximation and the actual value of f′(0.5)?

x 0 1

f(x) 1 2

A) 0

B) 0.176

C) 0.824

D) 1

Answer 1

y = 2^(x^2)

y = ln(2) (2^(x^2)) (2x)

Per the table the AROC is 1.

f'(1/2) = 0.82429555886596274569319955223376

so the difference is option B

Answer 2

One would approximate the value of f'(0.5) using the slope of a line between the two points, and the slope would be 1.

However, the actual value is given by taking the derivative of f(x) = 2^{x^2}, as shown:

d/dx[2x²]

=ln(2)⋅2x²⋅d/dx[x²]

=ln(2)⋅2x²⋅2x

=ln(2)⋅x⋅2^{x^2+1}

Thus the actual value of f'(0.5) = ln(2) ⋅0.5 ⋅2^{0.5^2+1} = 0.824, and the difference between the actual and estimate would be 1-0.824 = 0.176, answer B.

Answer 3

f'(x) is the instantaneous rate of change.

We can approximate the rate of change

by finding the slope between two, nearby points.

Slope = (change in y) / (change in x)

Slope = (2 - 1) / (1 - 0) = 1/1 = 1

So, f'(0.5) ~ 1

Our approximation is based on two nearby points.

If the nearby points were closer to x = 0.5

then our approximation would be more accurate.

In fact, the derivative is the slope, calculated when

the change in x approaches zero.