I have the derivative and the second derivative, now how do I find the actual function?

there is a lot of redundant information here. You need only the g '(x) and one of the g values. g(0) would be simplest to use. Multiply the brackets of g'(x) then integrate term by term and include a constant c. You can then find c using g=50 when x=0

g'(x) = (-2x^2-8x -8)(x-4) = -2x^3-8x^2-8x +8x^2 +32x+32 = -2x^3 +24x +32 g(x) = (-x^4)/2 + 12x^2 +32x + C ......C= a constant g(0) = C = 50 g(2) = -8 +48 +64 +50 = 154 g(4) = -128 +192 +128 +50 = 242 g(-2) = -8 +48 -64 +50 =26 (not 25) so g(x) = (-x^4)/2 +12x^2 +32x +50 all checks work except g(-2) = 26 I can't see any mistake, so can you check the given data? Is g(-2) = 26 ?

Expand the bracket g'(x) = -2(x + 2)^2(x - 4) g'(x) = -2(x^2 + 4x + 4)(x - 4) g'(x) = -2(x^3 + 4x^2 +4x - 4x^2 - 16x -16) g'(x) = -2(x^3 - 12x -16) g'(x) = -2x^3 + 24x +32 Now integrate! g(x) = -(x^4)/2 + 12x^2 + 32x + c *My formula is incorrect, this is the correct forumula but I can't figure out where I've gone wrong = -1/2 x^4 + 4/3 x^3 + 4x^2 + 32x + C But when x = 0, g(0) = 50 so c = 50

if given g ' = -2(x+2)^2(x-4) = - 2x^3 + 4x^2 + 8x + 32 g(x) = ∫ (-2x^3 + 4x^2 + 8x + 32) dx = = -1/2 x^4 + 4/3 x^3 + 4x^2 + 32x + C if g(0) = 50 ==> C = 50 g(x) = -1/2 x^4 + 4/3 x^3 + 4x^2 + 32x + 50

There is a ton of extraneous information in this problem. All you need is the first derivative and one of the given points. Rearranging g'(x) we can show that g'(x)=-2(x^3-12x-16). Taking the indefinite integral of g'(x), we get g(x)=(-1/2)x^4+12x^2+32x+C. Then plug in one of the points. For simplicity, I used g(0)=50. Then we have 50=(-1/2)(0)^4+12(0)^2+32(0)+C. Solving for C we get 50. So g(x)=(-1/2)x^4+12x^2+32x+50. You can plug in the other points to verify that this is correct.

g'(x) = (-2x^2-8x -8)(x-4) = -2x^3-8x^2-8x +8x^2 +32x+32 = -2x^3 +24x +32 g(x) = (-x^4)/2 + 12x^2 +32x + C ......C= a constant g(0) = C = 50 g(2) = -8 +48 +64 +50 = 154 g(4) = -128 +192 +128 +50 = 242 g(-2) = -8 +48 -64 +50 =26 (not 25) so g(x) = (-x^4)/2 +12x^2 +32x +50 all checks work except g(-2) = 26 I can't see any mistake, so can you check the given data? Is g(-2) = 26 ?

There is a ton of extraneous information in this problem. All you need is the first derivative and one of the given points. Rearranging g'(x) we can show that g'(x)=-2(x^3-12x-16). Taking the indefinite integral of g'(x), we get g(x)=(-1/2)x^4+12x^2+32x+C. Then plug in one of the points. For simplicity, I used g(0)=50. Then we have 50=(-1/2)(0)^4+12(0)^2+32(0)+C. Solving for C we get 50. So g(x)=(-1/2)x^4+12x^2+32x+50. You can plug in the other points to verify that this is correct.

Expand the bracket g'(x) = -2(x + 2)^2(x - 4) g'(x) = -2(x^2 + 4x + 4)(x - 4) g'(x) = -2(x^3 + 4x^2 +4x - 4x^2 - 16x -16) g'(x) = -2(x^3 - 12x -16) g'(x) = -2x^3 + 24x +32 Now integrate! g(x) = -(x^4)/2 + 12x^2 + 32x + c *My formula is incorrect, this is the correct forumula but I can't figure out where I've gone wrong = -1/2 x^4 + 4/3 x^3 + 4x^2 + 32x + C But when x = 0, g(0) = 50 so c = 50

if given g ' = -2(x+2)^2(x-4) = - 2x^3 + 4x^2 + 8x + 32 g(x) = ∫ (-2x^3 + 4x^2 + 8x + 32) dx = = -1/2 x^4 + 4/3 x^3 + 4x^2 + 32x + C if g(0) = 50 ==> C = 50 g(x) = -1/2 x^4 + 4/3 x^3 + 4x^2 + 32x + 50