# On which interval(s) is the function y = x^3 + 3/2x^2 - 6x + 27 increasing?a) (-inf, 1) b) (-inf, -2); (1, inf) c) (-inf, inf) d) (-2, 1) e) (-inf, -1) ; (2, inf)

*print*Print*list*Cite

### 2 Answers

y= x^3 + (3/2)x^2 - 6x + 27

First we will find the first derivative.

==> y' = 3x^2 + 3x - 6

Now we will determine the zeros.

==> 3x^2 + 3x - 6 = 0

Divide by 3:

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

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

==> x= -2 and x= 1

Then the function changes direction at the points x=-2 and x= 1

Then we have the following increasing or decreasing intervals.

==> ( -inf, -2) ( -2, 1) and ( 1, inf)

==> The function increasing when y' > 0

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

==> x> -2 and x > 1

==> x > 1

OR

==> x < -2 and x < 1

==> x < -2

Then the function is increasing when x<-2 and x > 1

==> x = (-inf, -2) U (1, inf)

**Then the answer is: b) ( -inf,-2) ; ( 1, inf).**

We need to find the interval of the values of x where the function f(x) = x^3 + (3/2)x^2 + 6x + 27 is increasing.

The function is increasing if the value of the first derivative is positive. Here:

f’(x) = 3x^2 + (3/2)*2x + 6

=> 3x^2 + 3x + 6

3x^2 + 3x – 6 > 0

=> x^2 + x – 2 > 0

=> x^2 + 2x – x – 2 > 0

=> x(x + 2) – 1(x + 2) > 0

=> (x – 1)(x + 2) > 0

This is positive when either

- x – 1 > 0 and x + 2 > 0

=> x > 1 and x > -2 ,

x > 1 meets both the conditions.

- x – 1 < 0 and x + 2 < 0

=> x < 1 and x < -2

x < - 2 meets both the conditions.

**So the required value of x for which the function is increasing is (-inf. , -2) U (1 , inf.) or option b.**