# Differentiation Q: Find all points on the graph of the given function where the tangent line is horizontal. F(x)=(x+1)(x^2-x-2)

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The function f(x)=(x+1)(x^2-x-2). All points on the graph of f(x) have to be determined where the tangent is horizontal.

The slope of a line tangent to the graph of a function f(x) at any point x = c is equal to the value of f'(c).

A horizontal line has a slope equal to 0.

f'(x) = (x + 1)(2x - 1) + (x^2 - x - 2)

(x + 1)(2x - 1) + (x^2 - x - 2) = 0

=> 2x^2 + 2x - x - 1 + x^2 - x - 2 = 0

=> 3x^2 - 3 = 0

=> x = 1 and x = -1

**The graph of f(x)=(x+1)(x^2-x-2) has horizontal tangents at the points where x = 1 and x = -1.**

**`f(x) = (x+1).(x²-x-2)`**

**`f(x) = (x+1).(x²-2x+x-2)`**

**`f(x) = (x+1).(x(x-2)+1(x-2))`**

**`f(x) = (x+1).(x-2).(x+1)`**

**`f(x) = (x+1)².(x-2)`**

**Now differentiating wrt x on both sides and equating it to zero:**

**`f'(x) = 2(x+1).(x-2) + (x+1)² =0`**

**`f'(x) = 2(x+1).(x-2) + (x+1)² =0`**

**`2(x+1).(x-2) + (x+1)² =0`**

**`(x+1) [2(x-2) + 1] =0`**

**`(x+1) [2x - 4 + 1] =0`**

**`(x+1) [2x - 3] =0`**

So,

(x+1) =0 or 2x-3 =0

So,

`x= -1 or 3/2`

So, the points are:

`(-1 , 0)` and `(3/2 , -25/8 )`