1
$\begingroup$

I was solving this question in my assignment:

Question

I didn't have any idea to approach this , so i assumed that the first two are correct and came to the conclusion that the graph would be one of these:

Graph of function

From this , i got [C] and [D] correct too . But i cannot seem to understand why [A] and [B] should be correct always.

$\endgroup$
0
$\begingroup$

Consider a contradiction:

let f(x) be a increasing function

i.e. $f'(x)>0$ now we have 2 cases to satisfy the given condition

CASE 1 $f(x)>0$ and $f''(x)<0$ now assume this holds true thus $f(c)=k$ where k is positive number

now consider $f(c-w)$ such that $c-w<c$ and as $f'(x)>0$ therefore $f(c-w)<f(w)$ or $f(c-w)<k$ or $f(c-w)=k-p$

we can say there is a '$w$'[actually infinately many] such that $ p>k$ due to which $f(c-w)<0$ [which condraticts the assumption of f(x)>0 for all x] as the derivative is always positive and the function keeps decreasing towards the negative side and in fact increasingly decreasing towards negative side

CASE 2 $f(x)<0$ and $f''(x)>0$ now assume this holds true thus $f(c)=k$ where k is negative number

now consider $f(c+w)$ such that $c+w>c$ and as $f'(x)>0$ therefore $f(c+w)>f(w)$ or $f(c+w)>k$ or $f(c+w)=k+p$

we can say there is a '$w$'[actually infinately many] such that $ p>|k|$ due to which $f(c+w)>0$ [which condraticts the assumption of f(x)<0 for all x] as the derivative is always positive and the function keeps increasing and in fact increasingly increasing [`increases at a higher rate as value of x increases] as the double derivative is always assumed positive

similarly considering a decreasing function will NOT lead to a contradiction

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.