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Exactly one *real* solution, right?
take the derivative and see that the function is strictly increasing because the derivative is always positive
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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@Ahmad1 answer is correct, but the first part is unnecessary
a cubic always has a zero
how does the derivative show you that the function only intersect the x axis once?
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....