Does the function f(x) = x^3 + 27x^2- 54 have any –ve values.

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neela's profile pic

neela | High School Teacher | (Level 3) Valedictorian

Posted on

f(x) = x^3+27x^2 - 54.

To examine for negative values:

Solution:

f(x) = x^3+27x^2 -54 < 0 for x = 0 obviuosly.

f(-26) = (-26)^3+27*(-26)^2 -54 =  26^2*1 -54 > 0

f(-27) = (-27)^3+27*(-27)^2 -54 = -54.

Therfore f(x)  has a root between -26 and -27.

Therefore f(x) < 0 for  x < -27.

Similarlt f(1) = 1^3 +27*1^2-54  28-54 < 0

f(2) = 2^3+27(2^2)-54  = 62> 0.

So there is a root between 1 and 2.

Also f(-1.5) = (-1.5)^3 +27(-1.5)^2 - 54 = 3.375>0

f(-1.4) = (-1.4)^3+27(-1.4)^2-54 = -3.824 < 0.

So f(x) has a real root between -1.4 and -1.5

Let  these roots be x1  , x2 , x3 in increasing order.

Then f(x) = (x-x1)(x-x2)(x-x3). So when x < x1 all 3 factors are gative. So f(x) is negative.

When  x1 < x  < x2 ,   f(x) = (+ve)(-ve)(-ve) = -ve. So when x is between lower two roots f(x) positive.

When  x2 < x < x3,  f(x) = ( f(x) = (+v1) (+ve)(-ve) = -ve.

When  When x > x3 all factors are positive. So f(x)  > 0.

Thesefore f(x) negative when  x< x1 or x <  -26.9...

f(x) - ve for x2 < x  < x3, where x2 and x3 are the roots between (-1.5 and -1.4) and (1 and 2).

 

william1941's profile pic

william1941 | College Teacher | (Level 3) Valedictorian

Posted on

To determine if the function f(x) = x^3 + 27x^2- 54 has negative values, we could use random values of x and see what the value of f(x) is. Instead let’s use differentiation. We will try to find the minima of the function and if it is negative it means the function is negative for some values of x, else if the minima of the function is positive it only has positive values.

Now for f(x) = x^3 + 27x^2- 54

f’(x) = 3x^2+ 54x

Equating this to zero, 3x^2+ 54x =0

=> 3x(x + 18) =0

=> x = 0 and x= -18

f’’(x) = 6x + 54

For x = 0, f’’(x) = 54. As 54 is positive the minima is at x= 0,

Now f (0) = -54, therefore the function has negative values.

Therefore the given function has negative values also.

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