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# Is there a number that is exactly 1 more than its cube?

## If there is such a number, it satisfies the equation $x^{3}+1=x \Leftrightarrow x^{3}-x+1=0 .$ Let the left-hand side of this equation becalled $f(x) .$ Now $f(-2)=-5<0,$ and $f(-1)=1>0 .$ Note also that $f(x)$ is a polynomial, and thus continuous. So by theIntermediate Value Theorem, there is a number $c$ between -2 and -1 such that $f(c)=0,$ so that $c=c^{3}+1$.

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This is problem number sixty nine of this tour Calculus, eighth edition, section two point five. Their number. That is exactly one more than its cube. Another way of saying this. If we take the number two b x can one number equal can one number acts equal one more, then that name that number. Cute. So this is what the equation is that we're trying to solve for. And if we rearrange this equation setting equal to zero, we can treat this as finding the route of the function where F is equal to execute, plus one minus X. And we wantto prove that there is a route to this function because then that would prove that there is a solution to the question given. And we're going to use the intermediate value through to confirm that. And with a lot of trial in there, we can choose a certain into role. In this case, we're going to choose an general from negative to to To where at negative, too. The function evaluated at native to gives. Thank you, Terry. Plus three, which is negative. Five and and two. We get eight plus one minus two or seven. So What we see here is that in the interval, from negative to to to the function takes on all the values from negative five to seven. For the reason that the function f is continuous, we see that this function is a polynomial which is always continuous. Ah, for the domain of all real numbers and in the specific interval from negative to to to thing values, much must range between this negative five and the seven. And since eagles from negative too positive, it must take the value zero, which is what we are interested in. We have confirmed in this interval specifically, there exists a number. According to the interment media Valley theme, there exist a number. There exists an X value that makes the statement true that we can figure out index value for which the function F is equal to zero. So we have confirmed that there does indeed exist a number that is exactly one more and it's cute According to the intermediate value, Theo

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