The curve C with equation y=fx passes through the point (3, 14/2)
given that f'(x)=2x+3/xsquared
a) find f(x) (5 marks)
b) verify that f(-2)=5 (1 mark)
c) find an equation for the tangent to C at the point (-2,5) giving your answer in the form ax+by+c=0
f'=(2x+3)/x^2
f'=2/x + 3/x^2
f= int f'= 2lnx -3/x
sorry im really confused, which part of the questions is this answering? can i tell you my answers and you say if they're correct.
a) y=x squared -3x to the power -1 - 1/2
b)i did this :)
c) 13x-4y+6=0
a) To find f(x), we need to integrate f'(x) with respect to x.
Given f'(x) = 2x + (3/x^2), we can integrate both sides to obtain:
∫f'(x) dx = ∫(2x + 3/x^2) dx
Integrating the right side of the equation, we get:
f(x) = ∫(2x + 3/x^2) dx
Splitting the integral into two parts:
f(x) = ∫2x dx + ∫(3/x^2) dx
Integrating each part separately:
f(x) = x^2 + (-3/x) + C
where C is the constant of integration.
b) To verify that f(-2) = 5, we substitute x = -2 into the expression for f(x):
f(x) = x^2 - (3/x) + C
f(-2) = (-2)^2 - (3/(-2)) + C
= 4 + 3/2 + C
= 8/2 + 3/2 + C
= 11/2 + C
We are given that f(-2) = 5, so we can solve for the constant C:
5 = 11/2 + C
Subtracting 11/2 from both sides:
5 - 11/2 = C
10/2 - 11/2 = C
-1/2 = C
Hence, the value of the constant C is -1/2.
c) To find an equation for the tangent to curve C at the point (-2, 5), we need to determine the derivative of f(x) at that point. Then we can write the equation of a line in the point-slope form: y - y1 = m(x - x1), where (x1, y1) is the given point and m is the slope.
First, we calculate the derivative of f(x):
f'(x) = 2x + (3/x^2)
Next, we substitute x = -2 into f'(x) to find the slope at (-2, 5):
m = f'(-2) = 2(-2) + (3/(-2)^2) = -4 + 3/(2^2) = -4 + 3/4 = -4 + 3/4 = -13/4
Using the point-slope form with the point (-2, 5) and slope -13/4:
y - 5 = (-13/4)(x - (-2))
y - 5 = (-13/4)(x + 2)
Multiplying through by -4 to clear the fraction:
-4(y - 5) = -13(x + 2)
-4y + 20 = -13x - 26
Rearranging and simplifying:
13x + 4y + 46 = 0
So, the equation of the tangent to curve C at the point (-2, 5) is 13x + 4y + 46 = 0.