Wiki of The Day: Mean-Value Theorem

Calculus Level 2

Given that $f(x)=\dfrac 13 x^3+x^2+2x+2020$ . How many real roots does $f(x)$ have?

2 0 3 1

1 solution

Chew-Seong Cheong
Oct 1, 2020

Given that $f(x) = \dfrac {x^3}3+x^2+2x + 2020$ , then $\dfrac {df(x)}{dx} = x^2 + 2x + 2 = (x+1)^2 + 1 > 0$ . This means that $f(x)$ is increasing for all $x$ and it can be at most one real root. Since $f(0)>0$ and $f(-1)<0$ , there is $\boxed 1$ real root between $-1<x<0$ .

@Lingga Musroji , you have to mention real in the problem statement because $f(x)$ has three roots though the other two are not real. A real root is a complex root, while a complex root may not be real. May just write "roots" instead of "root(s)" because it is understood. Grammatically it should be "many roots".

Chew-Seong Cheong - 8 months, 2 weeks ago

I already asked for "real root(s)"

Lingga Musroji - 8 months, 2 weeks ago

Nope, I added the word "real" for you. Also added "that". I am a moderator I can change the question. Now I bold real for you.

Chew-Seong Cheong - 8 months, 2 weeks ago

You can also argue that, as a cubic, it must have at least one real root. Though this seems to be the intermediate value theorem, not the mean value theorem - have I missed the connection with the title?

Chris Lewis - 8 months, 2 weeks ago

The MVT is what is used to prove that a function with strictly positive derivative is strictly increasing. It IVT tells us that a cubic has at least one real root. The MVT then tells us that there is exactly one. (Although I would probably use Rolle's Theorem, a special case of the MVT, to show that there cannot be two roots, since the derivative of $f$ is never zero).

Mark Hennings - 8 months, 1 week ago

Wiki of The Day: Mean-Value Theorem

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