# Find the function f(x)=x^3+ax^2+bx+c if f(x) has critical points at x= -2/3 and x= 5f(1)= 3/2

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### 1 Answer

You need to remember that the x coordinate of critical point of f(x) is one zero of derivative of f(x).

Hence, you need to find the first derivative such that:

`f'(x) = 3x^2 + 2ax + b`

You need to plug `x = -2/3` in f'(x) such that:

`f'(-2/3) = 3*4/9 - (4a)/3 + b =gt 0 = 4/3 - a*(4/3) + b`

You need to plug x = 5 in f'(x) such that:

`f(5) = 75 + 10a + b =gt 0 =75 + 10a + b`

You need to use elimination to solve the system of simultaneous equations such that:

`75 + 10a + b - 4/3+ a*(4/3)- b = 0`

Reducing b yields:

`75 - 4/3 + 10a + (4a)/3 = 0`

You need to multiply by 3 both sides:

`225 - 4 + 30a + 4a = 0 =gt 221 + 34a = 0 =gt 34a = -221=gta = -6.5`

`75-65 + b = 0 =gt 10 + b = 0 =gt b = -10`

You need to use the third condition,`f(1) = 3/2` , to find c such that:

`f(1) = 1- 6.5 - 10 + c =gt 3/2 = -15.5 + c =gt 3 = -31 + 2c`

`2c = 31 + 3 =gt 2c = 34 =gt c = 17`

**Hence, evaluating the coefficients of function yields:`f(x) = x^3 - 6.5x^2 - 10x + 17.` **