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The third last paragraph states that since the roots are simple (not repeated), it means that the polynomial f(x) changes sign on either side of the root.
However, it turns out that although true, some instructors do not find this as a sufficient explanation. In this case, we have to test each interval.
The real number line has been divided into three intervals.
`(-infty,-3)` , `(-3,3)` and `(3,infty)` from the roots -3 and 3.
We need to take any x-value from each of these intervals and substitute it into f(x). If that value is positive, then the function is positive for the whole interval. Conversely, if it is negative, then the function is negative for the whole interval.
Let's pick x=-4 for the first interval
Then `f(-4)=(-4)^4-81=175>0` , so the function is positive in the first interval. We want the function negative, so this interval is no good for us.
We already picked x=0 in the answer above, and that turned out to be negative which is what we wanted, so the interval (-3,3) is a solution.
Let's pick x=4 for the third interval.
Then `f(4)=4^4-81=175>0` , so the function is positive in the first interval. We want the function negative, so this interval is no good for us.
The inequality has solution on the interval (-3,3).
The inequality `x^4>81` is solved in a similar way to the equation `x^4=81` , except we need to switch the inequality every time we multiply or divide by -1.
`x^4=81` can be solved by factoring `f(x)=x^4-81` using a difference of squares.
`=(x^2-9)(x^2+9)` use difference of squares
`=(x-3)(x+3)(x^2+9)` the last factor has complex zeros, so can be ignored for inequalities
The points of interest for the inequality are x=3 and x=-3. Since each of the points are simple roots, this means that the polynomial `f(x)` changes sign on either side of the roots.
Since `f(0)=-81<0` , then the inequalilty is true on the interval `(-3,3)` and is false on the intervals on either side.
This means the inequality has solution on the interval `(-3,3)` .
Thank you for your help. much appreciated
need help creating the intervals and testing each
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