Question

Use the following graph to estimate the rate of change of the function at x=0.6 using the points (0,0) and (1,−0.5)

Answer 1

The rate of change of the function at x = 0.6 is approximately -0.72.

The graph provides the following information:

(0, 0) and (1, -0.5) points

We know that the graph represents a cubic function, so

f(x) = ax3+bx2+cx + d

At (0, 0) and (1, -0.5) = 0, f'(x) provides;

3ax² + 2bx + c = 0

∴ c = 0

Also

ax³ + bx² + cx + d = 0

d = 0

We now have

3a(1)² + 2b(1) + 0 = 0

3·a + 2·b = 0..............(1)

a(1)³ + b(1)² + c(1) + d = -0.5

a + b + 0×(1) + 0 = -0.5

a + b = -0.5..........(2)

When equation (2) is multiplied by 2 and subtracted from equation (1), the result is;

2 × (a + b = -0.5) = 2·a + 2·b = -1

3·a + 2·b - (2·a + 2·b) = 0 - (-1) = 0 + 1 = 1

a = 1

From

We have a + b = -0.5;

1 + b = -0.5 = 0

b = -0.5 - 1 = -1.5

The formula becomes

f(x) = x³ - 1.5·x²

The function's rate of change when x = 0.6 is thus stated as follows;

f'(x) = 3 × x² - 2 × (1.5)·x = 3·x² - 3·x

We have at x = 0.6;

f'(0.6) = 3·(0.6)² - 3·(0.6)

f'(0.6) = 3×0.6² - 3×0.6 = -0.72

The function's rate of change at x = 0.6 ≈ -0.72.

Answer 2

The rate of change of the function at x = 0.6 is approximately -0.72

Explanation:

The given information from the graph are;

Points (0, 0) and (1, -0.5)

We have that the graph is that of a cubic function, therefore;

f(x) = ax³ + bx² + cx + d

At f'(x) at (0, 0) and (1, -0.5) = 0 gives;

3ax² + 2bx + c = 0

∴ c = 0

Also

ax³ + bx² + cx + d = 0

d = 0

We have

3a(1)² + 2b(1) + 0 = 0

3·a + 2·b = 0..............(1)

a(1)³ + b(1)² + c(1) + d = -0.5

a + b + 0×(1) + 0 = -0.5

a + b = -0.5..........(2)

Multiplying equation (2) by 2 and subtracting it from equation (1) gives;

2 × (a + b = -0.5) = 2·a + 2·b = -1

3·a + 2·b - (2·a + 2·b) = 0 - (-1) = 0 + 1 = 1

a = 1

From

a + b = -0.5, we have;

1 + b = -0.5 = 0

b = -0.5 - 1 = -1.5

The equation becomes

f(x) = x³ - 1.5·x²

The rate of change of the function at x = 0.6 is therefore given as follows;

f'(x) = 3 × x² - 2 × (1.5)·x = 3·x² - 3·x

At x = 0.6, we have;

f'(0.6) = 3·(0.6)² - 3·(0.6)

f'(0.6) = 3×0.6² - 3×0.6 = -0.72

The rate of change of the function at x = 0.6 ≈ -0.72