Hostile integration find f(x) question
"The curve meets the x-axis at the origin and at the point A.
Given that f ′(x) = 3x1/2 − 4x−1/2 ,
(a) find f(x),
(b) find the coordinates of A.
I integrated the curve for part (a) but am just let with an equation with no coordinate values to sub in to find C. If you use the coordinates of the origin, you'd just get C=0 and f(x) = 0 but then you wouldn't be able to do part (b) ?!
Given that f ′(x) = 3x1/2 − 4x−1/2 ,
(a) find f(x),
(b) find the coordinates of A.
I integrated the curve for part (a) but am just let with an equation with no coordinate values to sub in to find C. If you use the coordinates of the origin, you'd just get C=0 and f(x) = 0 but then you wouldn't be able to do part (b) ?!
Original post by jessyjellytot14
"The curve meets the x-axis at the origin and at the point A.
Given that f ′(x) = 3x1/2 − 4x−1/2 ,
(a) find f(x),
(b) find the coordinates of A.
I integrated the curve for part (a) but am just let with an equation with no coordinate values to sub in to find C. If you use the coordinates of the origin, you'd just get C=0 and f(x) = 0 but then you wouldn't be able to do part (b) ?!
Given that f ′(x) = 3x1/2 − 4x−1/2 ,
(a) find f(x),
(b) find the coordinates of A.
I integrated the curve for part (a) but am just let with an equation with no coordinate values to sub in to find C. If you use the coordinates of the origin, you'd just get C=0 and f(x) = 0 but then you wouldn't be able to do part (b) ?!
Huh? That's exactly what you should get.
You get c=0 so that and you can then factorise that to get one intersection at the origin and one intersection at some other value of which I will let you find.
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