Dec 9, 2011 · Assume that f is a differentiable function such that f(0)=f'(0)=0 and f''(0)>0. Argue that there exists a positive constant a>0 such that f(x)>0 for all x in the interval (0,a). Can anything be concluded about f(x) for negative x's? Homework Equations The Attempt at a Solution I think I should use the MVT so here is what I tried:

Apr 4, 2012 · If you know about exterior derivatives, then this identity is equivalent to [itex]d^2 = 0[/itex]. To give a "physical picture" for this identity, first use Gauss's Theorem for [itex]\mathrm{div}(\mathrm{curl} \, \vec{F})[/itex] to get:

Feb 3, 2009 · If f(x_0)=c and c > 0 then the function must go up to atleast c, and since it is continuous it does not "jump" there. I was thinking of making a small partition of a,x_0, and b. I then know what the sup or inf values of the two rectangles (speaking of the hieghts) will be greater than or equal to x_0 for the upper sum.

Dec 15, 2024 · ##F_a=\frac 12(f_0\frac{c+v}{c-v}-f_0)=f_0\frac{v}{c-v}## in agreement with the official answer. The ‘beat frequency’ is in fact ##2F_a## corresponding to your answer. That’s where the confusion arose. You need to think about why this is so! Minor edit.

Mar 4, 2025 · My solution: Let ##f(x+y)=f(x)f(y)-[f(x)+f(y)]-1/2## İf ##y=x## we find functional equation that given us. So for ##x=y=1## then ##f(2)=-1/2## İf we evaluate ##x=1, y=2## at above equation ##f(3)=-3## My question is: What is the solution of that functional equation; I mean are there other...

Feb 22, 2025 · In your parabola example, if p > 0 and q > 0, the graph of ##f(x) = a(x + p)^2 + q## will be translations to the left by p and up by p of the graph of ##y = ax^2## Yes - it's intended that a is a real number but not necessarily positive.

SOLUTION: Given f(x) = x2 + 3x – 5, find f(0). -5 Did I do this right. Can you show me the steps

Mar 6, 2025 · ##M a_{cm} = F_{ext}## Hi, I started this thread to address the solution of this problem. ProblemView attachment 358160 A yo-yo of radii ##R_1=R## and ##R_2=\frac{7}{5}R## is acted upon by forces ##F## and ##\kappa F~~~(0<\kappa<\infty)## as shown in the figure on the right. The yo-yo rolls without slipping on the horizontal surface.

Jan 21, 2025 · From ##f(0)f(x) = f(x)## and with ##f(0)=0##, one can deduce that ##f(x) \equiv 0\, \forall\, x##. You can however rule that other solution out due to the fact that ##f(5) = 32##. You may need to add the requirement that ##f(x)## is continuous for all real ##x## to prove the already-found function is unique.

That's the same problem as Find the equation of the line that passes through the two points (0,2) and (2,4) except that when you finish you write f(x) instead of y. Use the slope formula: Use the point slope formula: Change y to f(x) f(x) = x + 2 Edwin