integration help (1 Viewer)

[sssona09]

help pls

a) if d^2y/d^2x = 6x+4 and dy/dx=8 and y=2 when x=1, find your site a function of x

b) if f''(x)=x, f'(2)=1 and f (3)=0, find f (x)

c) if f''(x) =-2x and the curve y=f (x) has a stationary point at (2,6), find the equation of this curve

thank uu

 

[JMP lova]

Could you clarify the meaning of ‘the site of function of x’, for ‘a)’ ?

 

[sssona09]

oh sorry - I meant "find y as a function of x"

 

[darkk_blu]

For a)
d^2y/d^2x = 6x+4
dy/dx = 3x^2+4x+c1
Since dy/dx=8 when x=1
8 = 3(1)^2+4(1)+c1
c1 = -5
dy/dx = 3x^2+4x-5
y = x^3+2x^2-5x+c2
Since y=2 when x=1
2 = (1)^3+2(1)^2-5(1)+c2
c2 = 4
Therefore y = x^3+2x^2-5x+4

 

[darkk_blu]

For b)
f''(x)=x
f'(x)=x^2/2+c1
Since f'(2)=1
1=(2)^2/2+c1
c1=-1
f'(x)=x^2/2-1
f(x)=x^3/6-x+c2
Since f(3)=0
0=(3)^3/6-(3)+c2
c2=-3/2
Therefore f(x)=x^3/6-x-3/2

 

[sssona09]

thank you for the help!! <3

 

[fan96]

Generally with these types of questions you should write out the primitive function including any unknown constants, then use the points that are given to you to figure out what the constants are.

 

[darkk_blu]

Also "has a stationary point at (2,6)" implies that at x=2, y'=0 and at x=2, y=6