1. ## Solutions of functions

How would you approach these questions:

$1)\ Show\ that\ x^3+x-9=0\ has\ only\ one\ real\ solution.$

Would you differentiate it and then show that the derivative must be greater than 0 for all values of x. Then find a point, which gives a negative value of $f(x)$ and a point that gives a positive value of $f(x)$.

$2)\ How\ many\ real\ zeros\ does\ p(x)=x^3-12x^2+45x-51\ have?$

Would the answer always be 3. Could there be a case, where it is not 3.

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5. ## q1

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6. Originally Posted by acevipa
How would you approach these questions:

$1)\ Show\ that\ x^3+x-9=0\ has\ only\ one\ real\ solution.$

Would you differentiate it and then show that the derivative must be greater than 0 for all values of x. Then find a point, which gives a negative value of $f(x)$ and a point that gives a positive value of $f(x)$.

$2)\ How\ many\ real\ zeros\ does\ p(x)=x^3-12x^2+45x-51\ have?$

Would the answer always be 3. Could there be a case, where it is not 3.
For 1, I would do as you said - the derivative is $3x^2 + 1 > 0 \ \forall x \in \mathbb{R}$. Now all you need to do is find one point where f is negative and one where it is positive.

For 2, an example of a 3rd degree polynomial which doesn't have 3 real roots is $(x+i)(x-i)(x+1) = (x^2 + 1)(x + 1) = x^3 + x^2 + x + 1$, which has one real root = -1 and two complex roots - $\pm i$

In this example, differentiate the polynomial and find how many zeroes the derivative has (this is easier - it's just a quadratic in x..), and then how many extremal points. This should tell you how many zeroes the polynomial has (why?)