domain of sqrt{-x^3+5x^2+9x-45}

Question

 domain of sqrt{-x^3+5x^2+9x-45}

Answer 1

The number under the square-root must be zero of positive.  So we set the inequality.

 

-x³ + 5x² + 9x - 45 ≥ 0

 

 

Factor by grouping.

 

x²(5 - x) - 9(5 - x) ≥ 0

 

(x² - 9)(5 - x) ≥ 0

 

(x - 3)(x + 3)(5 - x) ≥ 0

 

 

Next, we can find the zeros of the inequality.  These are points on the x-axis.  The zeros are

 

x = -3        ,     x = 3    ,       x = 5

 

 

Now we look for intervals of x in which the function is on or above the axis.  Since we have a negative cubic function, our intervals of x will be

 

(-∞, -3]∪[3, 5]

Answer 2

{-x^3+5x^2+9x-45}

 

Rearrange the terms and factor the equation

 

-x³+9x +5x²-45

 

Multiply equation with -1

 

x³-9x -5x²+45

 

Take factor by grouping

 

x(x² - 9) -5(x² - 9)

(x² - 9)(x - 5)

(x + 3)(x - 3)(x - 5)

 

sqrt{(x + 3)(x - 3)(x - 5)}

 

sqrt of -ve number is  not real

 

x+3<0

x<-3

 

x-3<0

x<3

x-5 <0

x<5

 

excluded from domain.

 

So domain is x>=5

 

SO in Interval notation[5,∞)