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

To find the equation of the curve and answer the related questions, we need to follow the given information and apply some calculus concepts. Let's go step by step:

a) To find f(x), we need to first integrate f'(x):

f'(x) = 2x + 3/x^2

Integrating f'(x) with respect to x, we get:

f(x) = ∫(2x + 3/x^2) dx

To integrate this, we split it into two separate integrals:

f(x) = ∫2x dx + ∫3/x^2 dx

Integrating each term, we have:

f(x) = x^2 + (-3)/x + C

where C is the constant of integration.

Since the curve C passes through the point (3, 14/2), we can substitute these values into the equation for f(x):

14/2 = (3)^2 + (-3)/(3) + C

Simplifying this equation, we find:

7 = 9 - 1 + C

C = -1

Therefore, the equation for f(x) is:

f(x) = x^2 - 3/x - 1

b) To verify that f(-2) = 5, we substitute x = -2 into the equation for f(x):

f(-2) = (-2)^2 - 3/(-2) - 1

Simplifying this expression, we get:

f(-2) = 4 + 3/2 - 1

f(-2) = 8/2 + 3/2 - 2/2

f(-2) = 11/2 - 2/2

f(-2) = 9/2

Therefore, f(-2) is indeed equal to 5.

c) To find the equation of the tangent to the curve C at the point (-2, 5), we need to find the derivative of f(x) and use it to determine the slope of the tangent at that point.

First, let's find f'(x) by differentiating f(x):

f(x) = x^2 - 3/x - 1

f'(x) = d/dx (x^2 - 3/x - 1)

Applying the power rule and the quotient rule, we can find the derivative:

f'(x) = 2x + 3/x^2

Now, substitute x = -2 into f'(x) to find the slope at (-2, 5):

m = f'(-2) = 2(-2) + 3/(-2)^2

m = -4 + 3/4

m = -4 + 3/4

m = -4 + 3/4

m = -13/4

The slope of the tangent at (-2, 5) is -13/4.

Using the point-slope form of a linear equation (y - y1) = m(x - x1), where (x1, y1) is the point (-2, 5) and m is the slope, we can find the equation of the tangent:

(y - 5) = (-13/4)(x - (-2))

Simplifying this equation gives:

(y - 5) = (-13/4)(x + 2)

Multiply both sides by 4 to eliminate the fraction:

4(y - 5) = -13(x + 2)

Expanding and simplifying:

4y - 20 = -13x - 26

Rearranging the terms to obtain the desired form:

13x + 4y = -6

Thus, the equation of the tangent to C at the point (-2, 5) is 13x + 4y = -6.