Calculus

Dazed

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Jun 15, 2005
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1. Show that the equation x^4+2x^2-2=0 has exactly one solution in [0,1]

2. How many solutions does the equation sin^2x-3x=5 have? (give reasons)
 
Dazed said:
1. Show that the equation x^4+2x^2-2=0 has exactly one solution in [0,1]
1) If f(x) = x^4+2x^2-2
2) The Rule of Signs says there is One, and ONLY one, Positive Real Solution.
3) f(0) = -2 < 0
4) f(1) = 1 > 0
5) The Positive Real solution is in [0,1].

2. How many solutions does the equation sin^2x-3x=5 have? (give reasons)
1) Rewrite: [sin(x)]^2 = 3x + 5
2) [sin(x)]^2 has Range [-1,1]
3) If there are intersections, they must be on -1 <= 3x+5 <= 1 ==> -2 <= x <= -4/3
4) The derivative of [sin(x)]^2 - 3x - 5 < 0 on [-2,-4/3], so it's not coming back for another go.

You write the last step more formally.
 
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