f:R-R a 1-1 differentiable function.Show f is strictly increasing/decreasing?
364736
This statement can be proven by proving the contrapositive: If f:R -> R is neither strictly increasing nor strictly decreasing, then f is not a 1-1 differentiable function.\nIf f is differentiable, then f being neither strictly decreasing nor strictly increasing is equivalent to f being a constant function over some nondegenerate interval in its domain. Thus f cannot be 1-1.\nIf f is 1-1, then the condition implies that f is not continuous, and thus is not differentiable.
599545
No. Look at f(x) = |x|. It is everywhere continuous but not differentiable with respect to x at x=0.\n\nThis is a stupid question, and you're stupid for asking it.
1074093
The function has zeroes at -6 and 0.  It decreases to negative infinity as x decreases from -6. It increases to infinity as x increases from zero.  The only relative maximum is at -4.
764395
Hey, firstly I need this corrected: f(x)= Ix^2=5x+6I, because there should not be an equal sign after the power of 2.... I can't do this w/o knowing... if it is subtraction than I will tell you that from what I rembember about this stuff is that absolute value functions are continuous but not differentiable I think.... Because at the point, the f(c) equals the f(x) which means it is cont. but the derivative of f(x) approaching the vertex from the left side does not equal the derivative of the f(x) from the right side...whcih means it is NOT differentiable.... sorry I can't remember more buddy
781968
I'm not sure if your question meant f(x)=2x(x-1)(x-4) or f(x)=x(x-1)(x-4)^2, so I will answer both.\n\n1) f(x)=2x(x-1)(x-4)\n(0, 1) U (4, ∞)\n\n2) f(x)=x(x-1)(x-4)^2\n(-∞, 0) U (1, 4) U (4, ∞)
938777
1)  \nTo find increasing and decreasing of a function, you need to take the derivative:\n\nsqrt(4-x) can be re-written as (4-x)^(1/2)\n\nTaking the power rule and chain rule, you get (1/2)(4-x)^(-1/2)(-1)\n\nSimplifying, you get -1/(sqrt(4-x)).\n\nTo determine increasing or decreasing of a function, look at the derivative.  If the derivative is positive, the function is increasing and if the derivative if negative, the function is decreasing.  Looking from the derivative, you can conclude that the derivative is only defined for x < 4 because otherwise, you would have a negative square root which has no real value.  Thus, for all x < 4, you can conclude that the derivative is negative (try plugging in a value for x).  \n\nTherefore, the function is decreasing from (-infinity, 4)\n-------------------------------\n2)\nTake the derivative of the function:\n\nf(x) = x^2 - 4x\nf'(x) = 2x - 4\n\nSetting the derivative equal to zero locates your extreme values:\n\n2x - 4 = 0 ---------------> x = 2\n\nNow perform the number line test to see how the function behaves to the right and left of x = 2.  If you plug in a value to the left of 2, let's say x = 1, you get a negative derivative, thus the function is decreasing.  If you plug in a values to the right of 2, let's say x = 3, the derivative is positive.  Therefore, you can conclude from these findings that x = 2 is a minimum.  Solving for the actual point, plug in x = 2 into original function.  You get y = -4.\n\nSince the function extends to infinity in the y-direction for negative and positive infinty of x, (2, - 4) is an ABSOLUTE MINIMUM\n--------------------\n3)\nAgain, take the derivative:\n\nf(x) = x^3 + 5x^2 -13x + 1\nf'(x) = 3x^2 + 10x - 13\n\nNow set the derivative equal to zero to find extreme values:\n\n3x^2 + 10x - 13 = 0\n(x - 1)(3x + 13) = 0\n\nTherefore, you have extreme points at x = 1, -13/3\n\nNow do number line test like above to see if you get max or min.\n\nRemember, \nnegative->positive change = minimum\npositive->negative change = maximum\n\nx = 1\nto the left (like x = 0) ---------> negative\nto the right (like x = 2) -------> positive\n\nThus, at x = 1, you have a minimum.  The exact point is (1,-6)\n\nx = -13/3\nto the left (like x = -5) ---------> positive\nto the right (like x = -4) -------> negative\n\nThus, at x = -13/3, you have a maximum.  Solve for y to get the exact point.\n\nNow determine values at x = -+ infinity \n\nAt x = +infinity, y = +infinity\nAt x = -infinity, y = -infinity.\n\nTherefore, since at x = 1 and x= -13/3 give you y values that are not the extreme values, these are local extremes.  Thus, x = 1 is a local minimum and x = -13/3 is a local maximum\n-----------\n4)\n\nIf you plug in x = 0, you get a positive value thus it is increasing from -infinity to 1.  if you plug in x = 6, you get a positive value, thus the function is increasing from 5 to infinity.  If you plug in a value between 1 and 5 (like 3), you get a negative value, thus the function is decreasing from x = (1,5)\n--------------\n\nHope this helps
760867
f(x) = (x+3)(x^2 + 16)
819945
1.  the idea here is to take the limit for both pieces and show that they match.  This is relatively straight forward.  The limit as x approaches 1 from the left of x cubed is simply "1"     \nThe limit as x approaches 1 from the right of 3x-2 is 3(1)-2 also equals 1.  Since these two match, we have proven that "f is continuous at 1"\n\n2. The fact that f is continuous at 1 is INsufficient to imply that is is differentiable there\n\n3. the derivative from the right is simple 3 since the derivative of 3x-2 is 3.  The derivative from the left of 1 is also 3 since:\n\nf(x) = x^3 implies that f'(x) = 3x^2 and f'(1) = 3(1)^2 = 3 as well.  Since the derivative from the right equals teh one from teh left, the function is differntiable at x =1
694927
What are you trying to do?  You have just defined f(x).  You have not said what to do with it.
490567
=-1\n2(0)+3(0)-1=f(x)\n2(0)=0\n3(0)=0\nminus 1\nequals negative 1 or -1
472136
False. For example: F(x)=x^2 increases for x>0.\n\nF(x)=x^3 always increases.
475920
Homework?\n\nSuppose f is not identically 0. Then there is an interval where f stays away from 0. The g be the step function of this interval to get a contradiction.
866864
Just differentiate it. You need f'(x) < 0.\n\nBy the quotient rule, f'(x) = (2x(x^2 + 4) - (x^2)(2x))/(x^2+4)^2.\nNow, since the bottom is a square, thus positive, you just need the numerator to be < 0:\n2x(x^2 + 4) - x^2 * 2x < 0\n2x^3 + 8x - 2x^3 < 0\n8x < 0\nx < 0\nThus it is decreasing on (-∞,0).
491386
Just substitute x+1 into f(x)\n\nf(x+1) = [ 6(x+1)^3 ] - [ 3(x+1)^2 ] + 4(x+1) - 9
664810
Let f(z) = x+iy\n\nNow, When |z| = 1, f(z) is in R, this means y = 0\n\nBut z = x1+iy1\n\nFor f(x1+iy1) = x, the function must map every point in the complex plane to the real axis.\n\nNow on diffentiating f(z) w.r.t. z, we have z'f'(z).\n\nThis quantity is 0. Thus f(z) is a constant.
277316
let |f|=k (some constant).let u,v be the real & img parts of 'f'.\n            the we have |f|=u^2+v^2, u^2+v^2=k^2\n      differentiating partially,wrt to x&y ,and \n\n    USING CAUCHY-REI EQ'S FROM TWO RESULTING EQ'S \n\nu finally get     (u^2+v^2)(du/dx)=0.\n                 &  (u^2+v^2)(du/dy)=0. \nfrom above two eq's   we have \n                     k^2=u^2+v^2=0  implies u=0&v=0\n\n                  so 'f' is constant.
909269
Take the derivative of the original function. For your given example, if I remember my calculus correctly:\nf(x)=y\nf'(x)=1\nGraph the new equation.