Can a function share an interval with another, different function?

I’m not quite sure how to put this into words, but I’m really curious. Could you have two (defined, continuous non-piecewise) functions intersect “over an interval”? To put that another way, if you picked a continuous interval, say 0 to 1, and knew the value of a function f(x) for any x along that interval, … Read more

Find a function with certain requirements

I’m trying to find a function y=f(x) that can be described as follows: f(x)=g(x)+c/(x−xa). With f(x) I want to design a function with the following properties: f(0)=0; f(x) has a maximum at x1,x1>0; the maximum of f(x) is max f(x_2) = 0, \quad x_2>x_1; f(x) has a vertical asymptote at x=x_a, \quad x_a<0; f(x) has … Read more

BMO1 2009/10 Q5 functional equation: f(x)f(y)=f(x+y)+xyf(x)f(y) = f(x + y) + xy

Find all functions f, defined on the real numbers and taking real values, which satisfy the equation f(x)f(y)=f(x+y)+xy for all real numbers x and y. I worked out f(0)=1, and f(−1)f(1)=0, but then I hit a wall. Answer Setting y=0 gives f(x)f(0)=f(x). Since f cannot be identically zero, it follows that f(0)=1. Setting x=1,y=−1 then … Read more

Using the Mean Value Theorem, show that if f′(x)>0f'(x) > 0 ∀x∈(a,b)\forall x \in (a, b) then ff is increasing on (a,b)(a, b)

Using the Mean Value Theorem, show that if f′(x)>0 ∀x∈(a,b) then f is increasing on (a,b). The Mean Value Theorem states: a function f which is continuous on the closed interval [a,b] (1) and differentiable on the open interval (a,b) (2) has at least one value c:a<c<b where f′(c)=f(b)−f(a)b−a. Set f(x) to be some function … Read more

Showing this function is continuous $ f:(x,y)\mapsto x^2+y^2$

I have the following function: $$f:\Bbb R^2 \to \Bbb R,\quad f:(x,y)\mapsto x^2+y^2$$ I want to show that this function is continuous by showing that $f^{-1}((a,b))$ is an open set. How do I approach this? I have proved that the function is a metric already. But I can’t think of how elements are mapped for some … Read more

Prove that the solution of y′+y=arctan(ex),y(0)=2y’+y=\arctan(e^x), y(0)=2 admits horizontal asymptote.

Let us consider the Cauchy problem: y′+y=arctan(ex),    y(0)=2 Prove that the function y(x) admits horizontal asymptote without solving the problem. Answer I’ll prove a more general case, instead. If limx→+∞f(x)+f′(x)=l, then: \begin{align} &\lim_{x\rightarrow+\infty}f(x)=l\\ &\lim_{x\rightarrow+\infty}f'(x)=0 \end{align} Suppose we have \lim_{x\rightarrow+\infty}f(x)+f'(x)=l Then we have e^x(f(x)+f'(x))\approx_{x\rightarrow +\infty} le^x (e^xf(x))’\approx_{x\rightarrow +\infty} le^x=(le^x)’ Since x\rightarrow le^x is positive and its integral … Read more

Prove that $X \subset Y \implies f^{-1}(X) \subset f^{-1}(Y)$

Let E and F be two sets and $f: E \to F $ be a function, and $X, Y \subset F$. Prove that $X \subset Y \implies f^{-1}(X) \subset f^{-1}(Y)$ My answer: Let $y \in X$, then $f^{-1}(y) \in f^{-1}(X)$. Since $X \subset Y$, then $y \in Y$. If $y \in Y$, then $f^{-1}(y) \in … Read more

What is the difference between these two statements involving minimums of a function?

All values at which $f$ has a local minimum. and All local minimum values of $f$. Answer Yet another way to repeat the previous answers: if the point $(x,y)$ on the graph of $f$ corresponds to a local minimum, then one says that this local minimum value “is $y$” and that it “occurs at $x$”. … Read more

Explaining why the absolute value of an odd function is even.

For the following: If f(x) is an odd function, then |f(x)| is _____. I said even, because I graphed an odd function and then the absolute value of it and ended up with an even function. The answer is correct but then my professor says this as an explanation: h(x)=|f(x)| h(−x)=|f(−x)|=|−f(x)|=|f(x)|=h(x) He didn’t explain very … Read more