7x^3-x^2<9-63x

Question

Find the intervals of x where the statement is true and answer in interval notation.

Answer 1

Subtract 9 and add 63x on both sides of inequality.

 

7x³ - x² + 63x - 9 < 0

 

 

Now factor the left side of the inequality by grouping.

 

x²(7x - 1) + 9(7x - 1) < 0

 

 

(x² + 9)(7x - 1) < 0

 

 

 

What do we know about products.  Multiplying a negative by a positive gives us a negative.  So we can set inequalities for the factors using that fact.

 

x² + 9 < 0            and              7x - 1 >0

 

 

                            or

 

x² + 9 > 0             and              7x - 1 < 0

 

 

 

x² + 9 < 0   will never be a true statement for any value of x.  This is because a negative number squared is positive.  And when that positive number is added to a positive number, the result is positive.

 

So we have to go with these pairs of inequalities.

 

 

7x - 1 > 0             and              7x - 1 < 0

 

 

Solve for x from both inequalities.

 

7x > 1                    and              7x < 1

 

  x > 1/7                and                 x < 1/7

 

 

 

Now evaluate the inequality by plugging in  x=2/7  and  x=0 to validate the solutions.

 

If we plug in x=2/7, we will see that the left side is greater than the right side.  It does not make the statement true.

 

If we plug in x=0, the statement is true.

 

0 < 9

 

 

Intervals of x that will make the statement true is  (-∞, 1/7).

 

Answer 2

Hello Mae,

 

Let's try to figure out when 7x³-x² is less than 9 - 63x.

 

What I would like to do is subtract the right side from the left so that I get a polynomial to solve.

 

7x³-x² < 9 - 63x

 

add 63 x to both sides

 

7x³-x²+63x < 9

 

subtract 9

 

7x³-x²+63x-9<0

Now we can factor and find when our polynomial is negative. 
(7x-1)(x²+9)<0

notice that x²+9 will always be positive! that means that we can look at just 7x-1

 

7x-1<0

 

7x<1

 

x<1/7