A function f(x) = x^3-8x^2-x+8 . Is the value of f(x) less than 0 for any value of x.

3 Answers

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degeneratecircle | High School Teacher | (Level 2) Associate Educator

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For this problem, the factorization is more easily done as

`x^3-8x^2-x+8=x^2(x-8)-1(x-8)=(x^2-1)(x-8).`

If we use the fact that `x^2-1` is easily seen to be negative on `(-1,1)` and positive everywhere else except ` ``x=+-1` (see graph), then the cases become a bit easier to test.

For `x-8>0,` which is the same as `x>8,` we also want `x^2-1<0` , but this never happens when `x>8.` So we need `x-8<0,` or `x<8.` In this case, we should simultaneously have `x^2-1>0,` so our `x` values are all `x<8,` except for `-1<=x<=1.`

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justaguide | College Teacher | (Level 2) Distinguished Educator

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The function f(x) = x^3-8x^2-x+8. To determine the values of x for which f(x) < 0, the inequality x^3-8x^2-x+8 < 0 has to be solved.

x^3-8x^2-x+8<0

=> x^3 - x^2 - 7x^2 + 7x - 8x + 8 < 0

=> x^2(x - 1) - 7x(x - 1) - 8(x - 1) < 0

=> (x - 1)(x^2 - 7x - 8) < 0

=> (x - 1)(x^2 + x - 8x  - 8) < 0

=> (x - 1)(x(x + 1) - 8(x + 1)) < 0

=> (x - 1)(x + 1)(x - 8) < 0

Now (x - 1)(x + 1)(x - 8) < 0 in the following cases:

(x - 1) < 0, (x + 1)> 0 , (x - 8) > 0

=> x < 1, x > -1 and x > 8

This is not possible for any value of x.

(x - 1) > 0, (x + 1) < 0 , (x - 8) > 0

=> x > 1, x < -1 and x > 8

This also is not possible for any value of x.

(x - 1) > 0, (x + 1) > 0 , (x - 8) < 0

=> x > 1, x > -1 and x < 8

This is possible for `x in (1, 8)`

The value of f(x) = x^3-8x^2-x+8 is also less than 0 if all the factors are less than 0:

(x - 1) < 0, (x + 1)< 0 , (x - 8) < 0

=> x < 1, x < -1 and x <8

=> x < -1

=> `x in (-oo, -1)`

The value of `f(x) = x^3-8x^2-x+8<0` in the union of sets `(-oo, -1)U(1, 8)`

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aruv | High School Teacher | (Level 2) Valedictorian

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`f(x)=x^3-8x^2-x+8`

`=(x-1)(x^2-7x-8)`

f(x)<0  if

`1. x-1<0 and x^2-7x-8>0`

`=> x<1 and (x-8)(x+1)>0`

a.

`x<1, x>8 and x> -1`

It is not possible.

b.

`x<1,x<8 and x<-1`

`=> x<-1`

`2. x-1>0 and (x-8)(x+1)<0`

`=>x>1 and (x-8)(x+1)<0`

c.

`x>1 , x-8>0 and x+1<0`

`=> x>1,x>8 and x<-1`

It is not possible.

d.

`x>1 ,x-8<0 and x+1>0`

`=> x>1,x<8 and x> -1`

`=> x in (1,8)`

Thus f(x)<0 if  `x<-1 or x in (1,8)` i.e.

If  ` x in (-oo,-1)U(1,8)` then f(x)<0 .

`x_(mn)`