Question
solve the inequality please x^4-6561>0
So far I have come up with x^4>9^4. I guess what's throwing me is the answer has to be in interval notation and for the life of me, I can't remember what that is! (for me, it used to be a blond moment - now I'm gray with no excuse)
So far I have come up with x^4>9^4. I guess what's throwing me is the answer has to be in interval notation and for the life of me, I can't remember what that is! (for me, it used to be a blond moment - now I'm gray with no excuse)
Answers
How about factoring it?
x^4 - 6561 < 0
(x^2 + 81)(x^2-81) < 0
(x^2+81)(x+9)(x-9) < 0
critical values are 9 and -9 , (x^2+81 = 0 has no real solution)
So we know that the correspoinding graph of
y = x^4 - 6561 crosses only at -9 and +9 and get lost in the first and 2nd quadrants for large positive and negative values of x
So we know for x> 9 or x< -9 we get positive results
and for all values between -9 and +9 the result is negaive
so ....
x^4 - 6561 < 0
for x < -9 OR x > +9
I will let you write it in the "interval notation" that was taught to you.
(BTW, this interval notation is actually something new for me, 15 years ago when I retired we just used the notation I used above. To me it still makes the most sense)
x^4 - 6561 < 0
(x^2 + 81)(x^2-81) < 0
(x^2+81)(x+9)(x-9) < 0
critical values are 9 and -9 , (x^2+81 = 0 has no real solution)
So we know that the correspoinding graph of
y = x^4 - 6561 crosses only at -9 and +9 and get lost in the first and 2nd quadrants for large positive and negative values of x
So we know for x> 9 or x< -9 we get positive results
and for all values between -9 and +9 the result is negaive
so ....
x^4 - 6561 < 0
for x < -9 OR x > +9
I will let you write it in the "interval notation" that was taught to you.
(BTW, this interval notation is actually something new for me, 15 years ago when I retired we just used the notation I used above. To me it still makes the most sense)
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