Let f(x) = 2 - abs( 2x - 1).?
f(x) = 2 -|2x-1| f(3) = 2 - |2(3) -1| = 2 - |5| = -3 f(0) = 2 - |2(0) -1| = 2 - |-1| = 1 and f(3) - f(0)........-4 ▬▬▬▬ = ▬▬ ..3..-..0............3 Now if the mean value theorem could be satisfied, then there would be a c such that f'(c) = -4/3 Of course the information I posted above was the results you obtained already. Now to show that there is no c such that f'(c) = -4/3 differentiate f(x) Note that f(x) can be written as a 2 part function .............2- (2x -1) = 2-2x + 1 = 3 -2x if x> 1/2 f(x) = 1 if x= 1/2 ............2 - (1-2x) = 2 - 1 + 2x = 2x -1 if x< 1/2 so more simply .......... 3 -2x if x> 1/2 f'x) = 1 if x= 1/2 ...........2x -1 if x< 1/2 Therefore if we differentiate f(x) we find that ...........-2 if x> 1/2 f'(x) = does not exist if x= 1/2 ............2 if x< 1/2 thus f'(c) never equals -4/3
Let f(x) = x^2-2x-8/x-4?
Greetings, x^2 - 2x - 8 = (x - 4)(x + 2) f(x) = x^2-2x-8/x-4 = x + 2 for x not equal to 4 a) Now taking the limit x-> 4, x does not equal 4, so lim x-> 4 (x + 2) = 6 b) lim x-> 0 (x + 2) = 2 c) lim x-> 2 (x + 2) = 4 Regards
Let f : R--> R , where f (x) = 2^|x| - 2^-x . Then f (x) is one-one or many one?
Here, f(x) is many to one. because if i take the value of x=0,thenf(0) =2^0 -2^0 = 1 – 1 = 0.and for x = -1,f(-1) = 2^ |-1| - 2^-(-1) = 2^1 - 2^1 = 2 – 2 =0.So.for to different values of x those are 0 and -1. we are having same values of f(x). So, the function is many to one.of course , for any negative value of x, the function always returns 0.
Let [math]f(x) = x^4 + x^3 + x^2 +x + 1[/math]. How would one find the remainder when [math]f(x^5)[/math] is divided by [math]f(x)[/math]?
[math]f(x^5) = x^{20} + x^{15} + x^{10} + x^{5} + 1[/math][math]x^{20} - x^{15} + 2x^{15} - 2x^{10} + 3x^{10} - 3x^{5} + 4x^{5} - 4 + 5[/math][math]x^{5}(x^{15}+2x^{10}+3x^{5}+4) - (x^{15}+2x^{10}+3x^{5}+4) + 5[/math][math](x^{5} - 1)(x^{15} + 2x^{10} + 3x^{5} + 4) + 5[/math][math]f(x)(x - 1)(x^{15} + 2x^{10} + 3x^{5} + 4) + 5[/math]So the remainder when [math]f(x^5)[/math] is divided by [math]f(x)[/math] is [math]5[/math].
Slope of the secant line? Let f(x) = x^2-8x?
a) To find the slope of a line from (x1, y1) to (x2, y2) to compute: slope = (y2 - y1) / (x2 - x1) In your case you have the points (2, f(2)) and (9, f(9)). The first thing you have to do is find the y-values at these points. f(2) = 2^2 - 8*2 = 4 - 16 = -12 f(9) = 9^2 - 8*9 = 81 - 72 = 9 Now use the formula above to find the slope of the secant line between the points (2, -12) and (9, 9). slope = (y2 - y1) / (x2 - x1) slope = (9 - (-12)) / (9 - 2) slope = (21) / (7) slope = 3 b) Do the same thing for the second problem. f(4) = 4^2 - 8*4 = -16 f(4+h) = (4+h)^2 - 8(4+h) = h^2 + 8h + 16 - 32 - 8h = h^2 - 16 This time find the slope of the line between (4, -16) and (4 + h, h^2 - 16) slope = (y2 - y1) / (x2 - x1) slope = ((h^2 - 16) - (-16)) / ((4 + h) - 4) slope = (h^2 - 16 + 16) / h slope = (h^2) / h slope = h Hope this helps you!
Let f(x) = (1+b^2) x^2+2bx+1 and let m(b) be minimum value of f(x). As b varies, the range of m(b) is? Will anyone tell me how to solve such questions?
This is a very simple question and as suggested by Dharin Shah, first try it yourself before seeing this solution.Now, here f(x) = (1+b^2) x^2+2bx+1Rearranging the terms, f(x) = x^2 + (bx)^2 +2bx + 1=> f(x) = x^2 + (bx+1)^2For this function to take minimum value first derivative should be zero and second derivative should be positive.df(x)/dx = 2x +(2b^2) x + 2bIf df(x)/dx = 0, Solving for x, we getx = -b/(1+b^2)Also second derivative comes out to be positive independent of x. So this function has only minima.Now this minima value m(b) = f(x) at x = -b/(1+b^2)=> m(b) = [b^2/(1+b^2)^2] + [ 1 + b* { -b/(1+b^2 ) } ]^2=> m(b) = [b^2/(1+b^2)^2] + [ (1+ b^2 -b^2 )/ (1+b^2) ]^2=> m(b) = [b^2/(1+b^2)^2] + [1/(1+b^2)^2]=> m(b) = 1/(1+b^2)The variation of minima of the function with varying b is asIt has maximum value at b = 1 and the decreases with nearly inverse square rate to zero.
Let f(x) = (lne^(2x))/(x-1) for x>1. If g ism the inverse of f, then g'(3) =?
f(x) = ( ln e^( 2x ) ) / ( x - 1 ) ln e^ ( 2x ) = 2x so f ( x ) = 2x / ( x - 1 ) let f ( x ) = y = 2x / ( x - 1 ) y ( x - 1 ) = 2 x yx - y - 2x = 0 ( y - 2 ) x = y => x = y / ( y - 2 ) replace x by y and y by x to get inverse of f ( x ) y = x / (x - 2 ) g ( x ) = x / ( x - 2 ) g' ( x ) = ( x - 2 - x ) / ( x - 2 )^2 = -2 / ( x - 2 )^2 g ' ( 3 ) = -2 ( 3 - 2 )^2 = - 2
Let f be a function defined by f(x)= 2x+1, x<2 and 1/2x^2 + k, x>2?
a) If the expression is continuous at x = 2, then f(2)- = f(2)+! 2(2) + 1 = 2²/2 + k 5 = 2 + k k = 3 b) Differentiate the function and consider setting up this kind of expression f'(2)- = f'(2)+ f'(x) = 2 , x < 2 . . . . .x , x > 2 You should get x = 2 for the function to be differentiable for all x values. c) You don't get the continuous function since the function is continuous at x = 2 with k = 3! Hence, that function can't be continuous and differentiable where k = 5. Good luck!