Find all relative extrema and points of inflection for the following function...
h(X)= X^2+5X+4/ X-1
min=
max=
inflection points=
I will assume you meant:
h(x) = x^2 + 5x + 4/(x-1)
h ' (x) = 2x + 5 - 4/(x-1)^2
= 0 for max/min
2x + 5 = 4/(x-1)^2
(2x+5)(x^2 - 2x + 1) = 4
2x^3 - 4x^2 + 2x + 5x^2 - 10x + 5 = 4
2x^3 + x^2 - 8x + 1 = 0
hard to solve, Wolfram has this
http://www.wolframalpha.com/input/?i=2x%5E3+%2B+x%5E2+-+8x+%2B+1+%3D+0
if you meant it the way you typed it ...
h '(x) = 2x + 5 - 4/x^2 - 0
= 0
2x^3 + 5x^2 - 4 = 0
an equally messy cubic ....
http://www.wolframalpha.com/input/?i=2x%5E3+%2B+5x%5E2+-+4+%3D+0
are you expected to be able to solve a cubic ?
To find relative extrema and points of inflection for the function h(x) = (x^2 + 5x + 4) / (x - 1), we need to follow a step-by-step process.
1. Start by finding the first derivative of the function, h'(x), which will give us the critical points (where the derivative equals zero or is undefined).
2. Next, find the second derivative, h''(x), to determine the concavity of the function. The points where the second derivative changes sign are the inflection points.
Let's go through each step in detail:
Step 1: Finding the first derivative, h'(x):
To find h'(x), we need to apply the quotient rule. The quotient rule states that if we have a function f(x) = g(x) / h(x), the derivative f'(x) is given by:
f'(x) = (g'(x) * h(x) - g(x) * h'(x)) / (h(x))^2
Let's apply this rule to our function h(x):
h(x) = (x^2 + 5x + 4) / (x - 1)
Using the quotient rule:
h'(x) = [(2x + 5) * (x - 1) - (x^2 + 5x + 4) * 1] / (x - 1)^2
Simplifying the above expression:
h'(x) = (-x^2 + 3x + 9) / (x - 1)^2
Now we have the first derivative, h'(x).
Step 2: Finding the second derivative, h''(x):
To determine the concavity and find the inflection points, we need to find the second derivative h''(x). Differentiate h'(x) with respect to x.
h''(x) = [-2x(x - 1)^2 + (x - 1)(-x^2 + 3x + 9)(2)] / (x - 1)^4
Simplifying further:
h''(x) = (2x^3 - 9x^2 + 7x + 18) / (x - 1)^3
Now we have the second derivative, h''(x).
Next, we need to solve the critical points and determine if they are minima, maxima, or neither, by checking the pattern of the concavity. In addition, we will find the points of inflection by determining where the concavity changes.
To find the critical points, set h'(x) equal to zero and solve for x:
(-x^2 + 3x + 9) / (x - 1)^2 = 0
Notice that h'(x) is undefined at x = 1, as the denominator becomes zero. So x = 1 is a vertical asymptote and not a critical point.
Now, we solve for x by setting the numerator of h'(x) equal to zero:
-x^2 + 3x + 9 = 0
This is a quadratic equation. Factoring or applying the quadratic formula will give us the critical points.
Once you find the critical points, substitute them back into the original function, h(x), and the second derivative, h''(x), to determine if they are minima, maxima, or neither.
To find the points of inflection, find where the second derivative, h''(x), changes sign. These will be the x-values where the concavity changes from concave up to concave down or vice versa.
By going through these steps, you should be able to find the relative extrema (minima and maxima) and the inflection points for the given function h(x).