anonymous
  • anonymous
Prove that 5x^3+4x-2=0 has exactly one solution.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
Exactly one *real* solution, right?
anonymous
  • anonymous
take the derivative and see that the function is strictly increasing because the derivative is always positive
anonymous
  • anonymous
first we assume that f(x)=5x^3+4x-2 , now take the interval (0,1) f(0)<0 , f(1)>0 since the function is continous, then it must have at least one root in this interval by the intermediate value theorem. also we have:\[f \prime(x)=15x^2+4>0\] thus the function intersects the x axis only once i,e the function has only one solution

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anonymous
  • anonymous
@Ahmad1 answer is correct, but the first part is unnecessary a cubic always has a zero
anonymous
  • anonymous
how does the derivative show you that the function only intersect the x axis once?
phi
  • phi
The first derivative is a function that defines the slope of the curve at every point. in this case, the first derivative is 15*x^2+4 no matter what x value you test, the slope at that x is positive (y values increases as x increases) now if you know at some x value, the y value is negative... and at some bigger x value, the y value is positive, you know you crossed the x-axis. you know you only crossed once, because the curve *never* goes down. Once you pass the x-axis, the y value never decreases (slope is *always* positive) and of course, the curve continue up forever....
anonymous
  • anonymous
Oh that makes sense thank you.

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