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Integration/differentiation

So I'm not sure how to solve thos equation.
A curve with the equation y=f(x) passes through (4,9) given that
f’(x)= 3√x/2 - 9/(4√x) + 2
Where x is greater than 0.
1)* Find f (x) giving each term in its simplest form.
2)* point P lies on the curve. The normal to the curve at P is parallel to 2y+x=0
find the x coordinate of P.
Ive done some working as follows, but i dont think both parts are right:
f’(x)= 3√x/2 - 9/(4√x) + 2 = 1.5x^½ - 2.25x^(-½) + 2

So f(x) = ∫(1.5x^½ - 2.25x^(-½) + 2)dx

= (1.5/1.5)x^(1.5) – (2.25/(½))x^½ + 2x + c

= √x³ - 4.5√x + 2x + c

f(4) = 9 = √4³ - 4.5√4 + 2(4) + c = 8 – 9 + 8 + c = 7 + c
c=2
Reply 1
Avatar for Frstu
Frstu
OP
1
Original post by Frstu
So I'm not sure how to solve thos equation.
A curve with the equation y=f(x) passes through (4,9) given that
f’(x)= 3√x/2 - 9/(4√x) + 2
Where x is greater than 0.
1)* Find f (x) giving each term in its simplest form.
2)* point P lies on the curve. The normal to the curve at P is parallel to 2y+x=0
find the x coordinate of P.
Ive done some working as follows, but i dont think both parts are right:
f’(x)= 3√x/2 - 9/(4√x) + 2 = 1.5x^½ - 2.25x^(-½) + 2

So f(x) = ∫(1.5x^½ - 2.25x^(-½) + 2)dx

= (1.5/1.5)x^(1.5) – (2.25/(½))x^½ + 2x + c

= √x³ - 4.5√x + 2x + c

f(4) = 9 = √4³ - 4.5√4 + 2(4) + c = 8 – 9 + 8 + c = 7 + c
c=2

Here is the thread:smile:
Reply 2
Avatar for TeeEm
TeeEm
19
Original post by Frstu
Here is the thread:smile:


part 1 is correct

part 2

find the gradient of the normal by rearranging it to y = mx +c

the gradient function (given in the question) finds the gradient of the tangent

set it equal to the negative reciprocal of the gradient of the normal

solve the resulting equation to find the x coordinate of the point of normality.

if you need the y coordinate sub the x into the equation of the curve found in part 1.

all the best
Reply 3
Avatar for Frstu
Frstu
OP
1
Original post by TeeEm
part 1 is correct

part 2

find the gradient of the normal by rearranging it to y = mx +c

the gradient function (given in the question) finds the gradient of the tangent

set it equal to the negative reciprocal of the gradient of the normal

solve the resulting equation to find the x coordinate of the point of normality.

if you need the y coordinate sub the x into the equation of the curve found in part 1.

all the best


Belated thanks!

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