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4 votes
Find the

coordinates of the points on the graph of
ƒ(x) = ½ x³ − ¹⁄2x² − 8x + 7 where the gradient is 4.

User Fera
by
7.0k points

1 Answer

2 votes

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).

User Jota Santos
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8.7k points