Answer:
(-8/3, 19/27) and (3, -17/2).
Explanation:
To find the coordinates of the points on the graph of ƒ(x) = ½x³ − ¹⁄₂x² − 8x + 7 where the gradient is 4, we need to find the points where the derivative of ƒ(x) is equal to 4.
First, we need to find the derivative of ƒ(x):
ƒ'(x) = 3/2x² - x - 8
Next, we need to set ƒ'(x) = 4 and solve for x:
3/2x² - x - 8 = 4
3/2x² - x - 12 = 0
Multiplying both sides by 2 to eliminate the fraction:
3x² - 2x - 24 = 0
Factoring the quadratic equation:
(3x + 8)(x - 3) = 0
So x = -8/3 or x = 3.
Now we can find the corresponding y-coordinates:
When x = -8/3:
ƒ(-8/3) = 1/2(-8/3)³ - 1/2(-8/3)² - 8(-8/3) + 7 = 19/27
So one point on the graph with gradient 4 is (-8/3, 19/27).
When x = 3:
ƒ(3) = 1/2(3)³ - 1/2(3)² - 8(3) + 7 = -17/2
So another point on the graph with gradient 4 is (3, -17/2).
Therefore, the coordinates of the points on the graph of ƒ(x) = ½x³ − ¹⁄₂x² − 8x + 7 where the gradient is 4 are (-8/3, 19/27) and (3, -17/2).