# Thread: Curve Analysis w/Variables-what values of k, etc

1. ## Curve Analysis w/Variables-what values of k, etc

The first one I'm having problems with because k is a constant and I'm not used to finding that in derivatives. The second one I don't even know where to start.

How do you approach these problems which have a variable such as k, a b c, etc in terms of derivatives and curve analysis anyways? I couldn't find anything about them on the tutorial.

2. Originally Posted by SportfreundeKeaneKent
okay, k, a, b and c are just constants. you treat them as you would any number when finding the derivative or whatever.

(1) We find the maximum of a function by finding it's derivative and setting it equal to zero, so let's do that

f(x) = x + k/x
=> f ' (x) = 1 - k/(x^2)
for max, set f ' (x) = 0
=> 1 - k/(x^2) = 0
=> k = x^2
we want max at x = -2
so k = (-2)^2 = 4

3. Originally Posted by SportfreundeKeaneKent

Here's the second one.

the possibility of inflection points occur when the second derivative is 0, so let's find the points where the second derivative is 0

y = x^3 + bx^2 + c
=> y' = 3x^2 + 2bx
=> y'' = 6x + 2b

set y'' = 0
=> 6x + 2b = 0
=> 3x + b = 0
=> b = -3x
but x = -1 (since we are dealing with (-1,5)
so b = 3

so we have y = x^3 + 3x^2 + c
we know that (-1,5) is a point on the curve, so

5 = (-1)^3 + 3(-1)^2 + c
=> 5 = -1 + 3 + c
=> c = 5 + 1 - 3
=> c = 3

so our curve is y = x^3 + 3x^2 + 3

now, the slope of the tangent line is given by y'

y' = 3x^2 + 2bx = 3x^2 + 6x
at (-1,5)
y' = 3(-1)^2 + 6(-1) = 3 - 6 = -3

so the slope of our tangent line at (-1,5) is -3

now using m = -3, (x1,y1) = (-1,5), we have by the point-slope form:

y - y1 = m(x - x1)
=> y - 5 = -3(x + 1)
=> y = -3x - 3 + 5
=> y = -3x + 2 ..............the equation of the tangent line at P