If f(x) = √ 2x – 3   , find  f –¹ (x)?
953940
Hi again \n\nY = √2x – 3 \n2x – 3 = y ²\nX =( y ² + 3 )/ 2\nf –¹ (x)  =( y ² + 3 )/ 2  =>   f –¹ (x)  = (x ² + 3 )/ 2   &  x ≥ 0 \n\nGood Luck \n try to solve them yourslef again if you got wrong , am sure you can solve them correctly , if you just try your best
872722
F'(x) = √(5-x) - x/[2√(5-x)]\n\nF''(x) = -1/[√(5-x)] - 1/[2√(5-x)] + x/[4√(5-x)]³\n= - 3/[2√(5-x)] + x/[4√(5-x)³]
461122
Since f(x) is a polynomial of degree two in x,\nLet f(x) = ax^2 + bx + c, where a,b,c are constants.\n\nLet's find f(x) first.\n\nf(0) = 1 implies, c = 1.\n\nf(1) = 3 implies, a + b + c = 3\nor  a + b = 2 , or b = 2 - a  ...(1)\n\nf(-1) = 5 implies, a - b + c = 5\nor a - b = 4 ...(2)\n\nSubstituiting (1) into (2), we get,\n\na - [ 2 - a  ] = 4\n2a = 6 or a = 3.\nThus, b = 2 - a = 2 - 3 = -1.\n\nHence, f(x) = 3x^2 - x + 1\n\nNote that by long division or by the identity below,\n\n3x^2 - x + 1 = 3(x + 1)^2 - 7x - 2\n             \nWe get,\n\n[ 3x^2 - x + 1 ] /[1 + x]^2 = 3 - [7x + 2]/ [1 + x]^2   \n\nAnd that together with below,\n\n[7x + 2]/ [1 + x]^2  = 7 /(x + 1) - 5 /(x+1)^2\n\nGives,\n\n[ 3x^2 - x + 1 ] /[1 + x]^2 = 3 - 7 /(x + 1) + 5 /(x+1)^2\n  \nNow, we have an "integratable form".\n\n∫ [f(x) / (1 + x)^2] dx\n\n= ∫ [3x^2 - x + 1] / (1 + x)^2] dx\n\n= ∫ [ 3 - 7 /(x + 1) + 5 /(x+1)^2 ] dx\n \n= 3x - 7 log(x + 1) - (5/[x + 1]) + C\n\nwhere C is any arbitary constant.\n\nNote: Sometimes log(x + 1) can also be written as ln (x + 1)\n\nRemark. It is generally accepted that log and ln are both in base "e", the Euler number. If its in a different base, then usually its said or stated. For example logarithm of x in base 2 can be written as log_2(x).\nJust symbols :)\n\nRegards
773471
√2 . √6 \n---- x ----- =\n. 3 .... 22\n\n√12\n------- =\n66\n\n\n√(2² x 3)\n------------- =\n66\n\n2√3\n------ =\n66\n\n√3\n-----\n33
2302
(1) = x\n(2) = / \n(3) = x
618071
simplify to f(x)=(sqrt(x-2)(x-3))/(x-4)(x+5)\nthen sub in 1/x\nto get (sqrt(1/x -2)(1/x -3))/(1/x -4)(1/x +5)
735529
f(x)=-2x^4+6x+7\nf(-x)=-2(-x)^4+6(-x)+7\n=-2x^4-6x+7\nnether odd nor even
936579
your question seems to be incomplete.\nso far....\nwidth=x\nlength= 2x-1
559365
f(x)=x+3\nif x=0, f(0)=0+3=3\nif x=2, f(2)=2+3=5\nif x=-2, f(-2)=-2+3=1\n\nfor g(x)=2x+1\nif x=0, g(0)=2*0+1=1\nif x=2, g(2)=2*2+1=5\nif x=5, g(5)=2*5+1=11\n\nfor h(x.y)=2x+y\nif (x,y)=(0,0) h(0,0)=2*0+0=0\nif (x,y)=(2,1) h(2,1)=2*2+1=5\nif (x,y)=(1,3) h(1,3)=2*1+3=5
884668
f(x)=-4x^3+5x^2+8\nf(-3)=-4(-3)^3+5(-3)^2+8\n=108+45+8 is not equal to zero\nso x+3 is not a factor
886499
1). f(x) = (x-3)(x²+3x-5) +2. Multiply out to get your answer.\n2). f(1) =1, so we have to solve\nx³-8x²+17x-10 = 0.\nWe know x = 1 is a root of this equation, so x-1 is a\nfactor of the left-hand side. The other factor, obtained\nby long division or synthetic division is\nx²-7x+10.\nNow set this equal to 0 and solve:\n(x-5)(x-2) = 0,\nso the other 2 solutions are x = 5 and x = 2.
270471
is the question\ny = √x +3\nor y =√(x+3)where both x and 3 are under the root\nFor the first one the domain would be x≥0\nfor the second one the domain would be x≥-3\n\nP.S. The expression under the sqrt cannot be negative.
1041686
That's easy:   f'(x)=\n\n2*(x)^(3/2) - 3 - (x)^(1/2)\n--------------------------------\n6*(x)^(5/2)
819082
Let f(x) = x ³/ (x² + 2x - 15) \n\n= x - 2 + (19x -30)/(x² + 2x - 15) (by division)\n\n= x - 2 + (19x -30)/[(x - 3)(x + 5)]\n\nSo f(x) is undefined when x = -5, 3\n\nAlso lim x→±∞ f(x) = x - 2\n\nThus f(x) is asymptotic to g(x) = x - 2 as x →±∞ and asymptotic to x = -5 and x = 3
865303
1) If h(x) = f(g(x)), then g(x) = 4+3x and f(x) = x^5\n2) Assuming your function is 3sqrt(x^2 -8), then\ng(x) = x^2 - 8 and f(x) = 3sqrt(x)
881627
f(x)=x^3+5x^2+3x+1\nf'(x)=3x^2+10x+3\nsetting it to zero\n(3x+1)(x+3)=0\nx=-1/3 and x=-3\nf''(x)=6x+10\nat x=-1/3 f''(x)=+ve and so min\nat x=-3 f''(x)=-ve and so max
464622
Use any recursive algorithm.\n\nFor instance:\n\nf(x) = f(x) + 1