Proving f(x) has one real root

[Miike012]

The problem asked to prove by contradiction that the function has only one real root. They used Roll'es theorem and proved that f ' (x) =/= 0 then they concluded that because f ' (x) =/= 0 then by contradiction the function can have atleast one real root...


Look at the picture it has the answer to the solution...

Question:
I don't understand the logic... I made up a graph where f ' (x) = 0 but yet f(x) still only has one real root... so how is disproving that f'(x) can never equal zero prove that f(x) has atleast only one real root when I obviously showed a scenario in which f'= 0 but f only has one root

 

[vela]

The proof is saying because f'(x)>0 for all x, there is only one real root. You're confusing that statement with its converse, "f(x) has only one real root; therefore, f'(x)>0," which isn't generally true.

 

[Miike012]

thank you