Let \(\displaystyle{f{{\left({x}\right)}}}={\left\lbrace{x}^{{2}}+{10}{x}+{20}\right\rbrace}\) then the number of real

Samara Richard

Samara Richard

Answered question

2022-03-21

Let f(x)={x2+10x+20} then the number of real solutions of f(f(f(f(x))))=0.

Answer & Explanation

Melody Gamble

Melody Gamble

Beginner2022-03-22Added 10 answers

The minimum of f(x), a concave up quadratic, occurs at x=102(1)=5, and f(5)=5. Then the turning point of f(f(x)) will occur at f(f(5))=f(5)=5, and so on until f4(x) (which denotes f(f(f(f(x)))). There is only one minimum of f(x), so there is only one turning point of f4(x).
f4(x) is also concave up, since the leading term of f(x), f(f(x))... does not change sign, so the leading term is positive.

Do you have a similar question?

Recalculate according to your conditions!

Ask your question.
Get an expert answer.

Let our experts help you. Answer in as fast as 15 minutes.

Didn't find what you were looking for?