prove that for any real number k , the polynomial given by x^3+x+k has exactly one real root?

Answer 1

Let p(x) = x³ + x + k.
Since p(x) < 0 as x approaches -infinity.
and p(x) is positive for x large enough,
the intermediate value theorem tells us
that there is at least 1 real root.
Next, Rolle's theorem tells us that
there must be a real zero of p'(x) between any
2 zeros of p(x). But p'(x) = 3x²+1, which has no
real zeros. Consequently, p(x) cannot have
more than 1 real root. Thus it has exactly
one real root.

Answer 2

This polynomial must have at least 1 real solution (when x goes to +infinity it diverges positively and when it goes to -infinity it goes the other way so it must intersect the real axis because it is a continuous function.
It can only have either 1 or 3 solutions (distinct or not it does not matter. If it had 3 you could write it as

(x-a)*(x-b)*(x-c) but if you do that, do the multiplications and compare the terms it turns out the term multiplying x^2 must be zero, and that term is
a^2+b^2+c^2 (and k=-abc). Now, the sum of 2 or more real numbers, squared, can be zero only if the individual numbers are zero. It means that it cannot have 3 solutions, so it can only have one.

Answer 3

as the cubic function has always at least 1 real root
k here represent moving of the polynomial up & down , so the only way that this fucntion does not have another real root is that the function has no tops nor bottoms (i.e. min & max values)
so we can get f' (x)
f'(x) =3x^2 +1
put it f'(X)=0
3x^2 +1=0
3x^2 = -1
as there is no real square root for -1 , so there is no tops & bottoms to this curve so it will cut the x axis once only thruoghout all values of k
: )
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for cidyah
this is completely wrong
u mean if
a *b =20
so either a=20 or b=20
& it is obivous that 1*20,2* 10,4*5 can also be values of a& b,there is still fractions & irrationals
no this only in case of 0
a*b=0
so either a=0
or b=0

Answer 4

y = x^3+x+k
dy/dx = 3x^2 +1
for max & min => dy/dx = 0
=> 3x^2 +1 = 0
=> x^2 = -1/3
=> no real sol'ns
=> no max & min OR f(x) is increasing
=> cut X axis once

QED

d^2y/dx^2 = 6x
pt of inflection => d^2y/dx^2 = 0
=> 6x = 0
=> x= 0
=> y = k

pt of inflection = (0,k)

Answer 5

x^3+x=-k
x(x^2+1)=-k
x=-k
x^2+1 =-k
x^2=-(k+1)
square of a number is negative, so the roots are imaginary, not real.