What are the x values of all the critical points of this function?:?
409149
If you are looking for only the real values,\n\ndifferentiating once,\nd(x³-6x²+12x-4)/dx = 3x^2-12x+12\n\n3x^2-12x+12 = 0\n\nx^2 - 4x + 4 = 0\n\n(x-2)^2 = 0\n\nx = 2\n\nwill be the critical point;\n\ndifferentiating twice,\n\n2x-4=0\nx=2\n\n+ve constant, so that the function will have a minimum at this point.\n\n\nDaneengineer@yahoo.com
978117
The graph of f(x) is a line, which means it can't have an asymptote.\n\ng(x) = log5(2x) should have a vertical asymptote x = 0.
713860
1.Xc=1/2pifc\nc=1/2pifXc\n\n2.ib=ibq+rt(2is)\nib(1-q)=rt(2is)\nsquaring\n[ib(1-q)]^2=2is\ns=[ib(1-q)]^2/2i\n\n3.Q=Xl/R\nXl=Q/R\n\n4.Lt=L1+L2+2Lm\n2Lm=Lt-L1-L2\nLm=[Lt-L1-L2]/2\n\n5.Es=E1+E2+E3\nE2=[Es-E1-E3]\n\n6.R1=i2R^2/i2\ni2R^2=i2R1/i2\nR=rt[i2R1/i2]\n\n7.R1/Rx=R2/R3\nRx=R1R3/R2\n\n8.y=mx+b\nb=y-mx\n\n9.i2=i1R1/i1
179915
There will be a vertical asymptote at x=0. When x values get increasingly closer to zero from the right, the graph approaches infinity. When the x values approach 0 from the left, the graph approaches negative infinity.\n\nSimilarly, there is also a horizontal asymptote at y=4. As x approaches infinity, the graph approaches 4 (1/infinity + 4 = 4). Also, as x approaches negative infinity, the graph also approaches 4 (1/negative infinitive + 4 = 4).
885970
Domain: All real numbers\nNo vertical asymptotes\nHorizontal asymptote: lim_x->inf 1/(x+9/x) = 0\nCritical points:\ny' = [x^2+9-x(2x)]/(x^2+9)^2\n= [9-x^2]/(x^2+9)^2\nx = -3, 3
303836
Nope,\n\n   The log function blows up to infinity.  (If you don't believe this, give me a value for y=f(x) = K * log(x).  Ha ha! I can give you a value for x such that f(x) is greater than the value you gave me! )\n\n  Your pal from Calc I\n   cuntsucker
565951
the answer is zero b'coz at the end, X-X is equal to zero.
823012
Either one is possible. Consider y=x² (one critical value, which is not an inflection point) and y=x³ (one critical value, which IS an inflection point).
867304
f'(x)=-3+10x-3x^2\nf''(x)=10-6x\n\nmax and min are where f'(x)=0\n3x^2-10x+3=0\n(x-3)(3x-1)=0\nx=3,1/3\n\nto see if they are max or min use f''(x)\nf''(3)=-8   means concave down= max at x=3\nf''(1/3)=8   means concave up= min at x=1/3\n\ninflection points are where f''(x)=0\n10-6x=0\nx=5/3\nf''(0)=10 means concave up on the interval (-inf,5/3)\nf''(5)=-20  means concave down on the interval (5/3,inf)
745152
THE VALUE OF THE FUNCTION IS....-4(X^2+2X+1)+7\n                                                              -4(X+1)^2+7\n  HENCE THE VALUE ATTAINS A MIN WHEN (X+1)^2 IS MIN  AND THE MIN VALUE IS ZERO WHICH OCCURS AT X=-1 ...\nAND THE VALUE IS 7\nTHE MAXIMUM VALUE OCCURS WHEN X IS INFINITE..
735726
For x = -2, f(-2) = -8 - 4 = -12\nFor x = 1, f(1) = 4 - 8 = -4\nFor x = 2, f(2) = 8 - 8/4 = 6\nFor x = 3, f(3) = 12 - 8/9 = eleven and one-ninth\nFor x = 4, f(4) = 16 - 8/16 = fifteen and one-half\n\nAs x approaches 0 from either side, the value of the function approaches negative ∞.\n\nAs x gets very large, the graph approaches y = 4x from below as an asymptote. You can see that even by the time you get to x = 4, the graph gets pretty close to y = 4x.\n\nGraph those values. You don't need calculus.\n\nNice explanation by Wal. Openpsych has the value of f(-2) wrong.\n\nLet's take the first derivative of this and see where the curve turns.\n\nf'(x) = 4 + 16 / x^3\n0 = 4x^3 + 16\nx^3 = -4\nx = -cube root(4)\n\nThis is a local maximum. For x less than -cube root(4), the curve approaches x = 4y from below. For x greater than -cube root(4), the curve approaches negative  ∞.
879794
can you say what you mean by imageset of the function?\ndo youi mean pre image and image,namely the ordered pairs?\nif so\nx=4,y=0;x=1,y=-9;x=0,y=-12\nx=-5,y=-9,x=-2,y=-12\nso the ordered pairs are\n(4,0);(1,-9);(0,-12);(-5,-9);(-2,-12)
809962
1. f(x)= 3x^4 + 4x^3, -2 <_ x <_2\nf'(x)= 12x^3 + 12x^2 = 0\n12x^2(x+1)=0\nso the critical points are: x=0 or x=-1 \nf''(x)= 36x^2 +24x\nf''(0)=0, so this test is inconclusive.\nf''(-1) = 36 -24 >0 so -1 is a local minimum.\n\nNow, since you want an absolute max and min over the closed interval, you need to compare the values of f at the end points with the values of f at the critical points:\nf(-2) = 3(-2)^4 +4(-2)^3 = 16\nf(2) = 3(2)^4 +4(2)^3 = 80\nf(-1) = 3(-1)^4 +4(-1)^3 = -1\nf(0) = 0\nthe largest value happens when x=2, which means this is the abs. maximum over the interval\nand the smalles value happens when x=-1, which means this is the abs. min over the interval.\n\n\n2. f(x)=(x^2 + x)^2/3, [-2,3]\n\n3. f(x)= [6-4x], [-3,3]\n\n/
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
892270
First of all, you need to actually guess what the limit will be. \nf(x) = (x+2)(x-2) / (x+2) = x-2 as long as x is not equal to -2, so a good guess would be -4, since its very close to -4 when x is very close to -2.\n\nLet ε>0. Choose δ=ε. Then if 0 < |x- -2| = |x+2| < δ, we have:\nf(x) = (x+2)(x-2)/(x+2) = x-2, since x isn't equal to -2.\nSo |f(x) - -4| = |x-2- -4| = |x+2| < δ = ε, as required.
836562
f'(x)=2-5sinx\nsetting this to zero\nsinx=2/5\nabsolute extreme=2(2/5)+cosarcsin(2/5)
860585
Very ingenious to have a "That's Me" account. You can give yourself ten points over and over.\n\nfor x = -3, x^2 = 9 and 3x = -9. Result = 0\nfor x = -2, x^2 = 4 and 3x = -6. Result = -2\nfor x = -1, x^2 = 1 and 3x = -3. Result = -2\nfor x = 0, x^2 = 0 and 3x = 0. Result = 0\nfor x = 1, x^2 = 1 and 3x = 3. Result = 3\n\nOnly your zero value is correct.\n\nThis is an upward-opening parabola with its minimum point at \n\nx = -1.5\n\nIt crosses the x-axis at (-3, 0) and (0, 0). Obviously, it crosses the y-axis at (0, 0) also.