# how to find the particular equation of the cubic funtion? given local maximum(5,10) and point of inflection (3,2)

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You need to consider a cubic function `f(x)=ax^3+bx^2+cx+d.`

The problem provides you the information that the function reaches a local maximum at x=5, hence the root of equation `f'(x)=0` is `x=5 ` such that:

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

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

The problem provides you the information that the function has an inflection point at x=3, hence the root of equation `f''(x)=0 is x=3` such that:

`f''(x) = 6ax + 2b`

`f''(3) = 18a + 2b `

`18a + 2b = 0 =gt 9a + b = 0 =gt b = -9a`

You need to substitute `-9a` for `b` in `75a + 10b + c = 0` such that:

`75a + 10*(-9a) + c = 0 =gt -15a + c = 0 =gt c = 15a`

You need to remember that the local maximum and inflection point lie on the graph of the function such that:

`f(5)= 10 =gt 125a + 25b + 5c + d = 10`

You need to substitute `-9a` for `b` and `15a` for`c ` such that:

`125a - 225a + 75a + d = 10`

`-25a + d = 10`

`f(3) = 2 =gt 27a+9b+3c+d = 2`

You need to substitute `-9a` for `b` and `15a` for `c` such that:

`27a - 81a + 45a + d = 2`

`-9a + d = 2`

You need to subtract `-9a + d = 2` from `-25a + d = 10` such that:

`-25a + d + 9a - d= 10 - 2`

Collecting like terms yields:

`-16a = 8 =gt a = -1/2`

Substituting `-1/2` for a in `-9a + d = 2` yields `9/2+d=2 =gt d=2-9/2`

`d = -5/2`

Substituting `-1/2` for a in `b = -9a` yields:

`b = 9/2`

Substituting `-1/2` for a in `c = 15a` yields:

`c = -15/2`

Substituting `-1/2` for a, `9/2` for b, `-15/2` for c and `-5/2` for d in cubic function yields `f(x) = -x^3/2 + 9x^2/2 - 15x/2 - 5/2`

**Hence, evaluating the cubic function under given conditions yields `f(x) = -x^3/2 + 9x^2/2 - 15x/2 - 5/2` .**