calculus
Is there any continuous function $y=f(x)$ on $[0,1]$ that satisfies the following conditions?
$$
f(0) = f(1) = 0,
$$
and
$$
f(a)^2-f(2a)f(b)+f(b)^2<0,
$$
for some $0<a<b<1$.
I tried to test with several functions (with different $a,b$) but non of them satisfied.
Any help is appreciated. Thank you in advance.
Just try drawing what the graph could potentially be on paper, and find an equation for it. Here's a freebie for the first one (number 74): $$g(x)=\dfrac{1}{x-3}$$
Note: I am not taking calculus yet, but I do know about limits and such... I don't know if there is a faster way because my way seems inefficient and not algebraic-like.
EDIT: For the second one, maybe this would work:
$$f(x)=\begin{cases} -\frac{1}{x^2+\frac{1}{2}}, \ \ \ \ x < 0 \\ 0, \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ x=0 \\ \frac{1}{x^2+\frac{1}{2}}, \ \ \ \ \ \ \ \ x > 0 \\ \end{cases}$$
YET ANOTHER EDIT: Your answer for the final one (nonremovable discontinuity) is correct.
zeros of $f$ are values $c$ such that $f(c) = 0$.
I believe there is a problem with your graph at $x = -3$.
Your graph is not defined on $(2,\infty)$, so I don't believe it is continuous on $(0,\infty)$
This is what I am getting:
Those should be negative numbers on the left. Hehe.
Best Answer
What you want is a function that is highly peaked around $x=2a$. And you need to enforce the conditions that $f(0)=f(1)=0$; for that you can put in a factor of $x(1-x)$.
So try
$$ f(x) = x(1-x)e^{-20(x-\frac12)^2} $$ with $a=\frac14$ and $b=\frac34$.
There, $$f(a)^2 - f(2a)f(b) + f(b)^2 \approx -0.00766 <0$$