Need help solving this function:  Find f[g (x)]  f(x) = -5x + 7?
825459
what is the second function?
935264
Actually, the answer is *not* 4+x.\n(f/g)(x) = (16-x^2)/(4-x)\nNow, factorise the top:\n(4+x)(4-x)/(4-x).\nSo *if x is not 4* then (f/g)(x) = 4+x. However, the function is not defined at x = 4.
314638
7\n\ny-2x=5x\n\ny-2x+2x=5x+2x\n\ny=5x+2x\n\ny=7x\n\n7=slope
709230
1) because\nf(-5)=-5+7 =2
896061
f(x)=16-5x\nf(-3)=16-5(-3)\n=16+15=31
838764
- 7 - 4x < 13\n\n-7 - 4x + 7 < 13 + 7\n\n- 4x < 20\n\n- 4x/- 4 > 20 /- 4\n\nx > - 5\n\nThe answer is x > - 5\n\n- - - - -s-
195023
(7/x) - (8/5x) = ?\n\n[(5*7) - 8]/ 5x = ?\n\nAnswer: 27/5 x
751599
Primarily this is an expression raised to the fifth power, and so that is the outer function:\nf(x) = x^5,  or f(u) = u^5\n\nClearly, to get your function, you  need to replace x by\nx^2 + 4x + 6, and so that is the inner function, g(x).\nChecking, note that\nf(g(x)) = f(x^2 + 4x + 6)\n           = (x^2 + 4x + 6)^5\n\nI expect you know the "chain rule" for differentiating a composite function:\ndy/dx = (dy/du)*(du/dx)\n         = 5u^4*(2x + 4) and you finish this by replacing\nu with x^2 + 4x + 6 and taking out the common factor from \n2x + 4.
258077
X = -1\nf(-1) = 4*(1/7) = 4/7\n\nX = 1\nf(1) = 4*7 = 28\n\nX = 2\nf(2) = 4*7^2 = 4*49 = 196
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
447170
Change form to: \n\n5x^2 + 7x + 2 = 0 \n\nand use the quadratic formula. Or plot the graph of: \n\ny = 5x^2 + 7x + 2 \n\nand find the x-axis intersects. \n\nOr just do this: \n\n5x^2 + 7x + 2 = 0 \n\n5x^2 + 5x + 2x + 2 = 0 \n\n5x(x + 1) + 2(x + 1) = 0 \n\n(x + 1)(5x + 2) = 0 \n\nHence, \n\nx(1) = - 1 \n\nand \n\nx(2) = (-2)/5 \n\nsince one of the brackets has to equal zero to equal the left hand side. \n\nHope it's understandable and that it helped.
87883
First write y as a product instead of a quotient:\n\n(2+x)*(2-3x)^(-1)\n\nNow you should be able to do this. \nUse 2+x as your f and (2-3x)^(-1) as your g. \nYou will have to use the chain rule to find g'.
1161844
use this formula:\n\n[ g(h)  -  g(0)  ]    /   [ h  -  0 ]\n\ni.e.\n\n [  2/(h+1)   -   2/(0+1)  ]  / [ h  -  0  ]  =  [ 2/(h+1)   -  2]   /h\n\n=  [( 2 -2h  -2) / (h+1) ]  / h   =  -2/(h+1)\n\nTry it for f(t)
1087517
x-(-2 -7)=-1\nx+2x+14=-1\n3x=-15\n\nx = -5\n\ny=-x-7\ny=-(-5) - 7\ny=5-7\n\ny = -2
490567
=-1\n2(0)+3(0)-1=f(x)\n2(0)=0\n3(0)=0\nminus 1\nequals negative 1 or -1
635593
g(x) = 5x - 4\n\nWith x = 2a, we have:\ng(2a) = 5(2a) - 4\ng(2a) = 10a - 4
772631
Assuming this one is like your other questions, then\nAssuming h(x) = f(g(x)), then \ng(x) = x-1 and f(x) = 2x^5/3 + 5x^2/3