# Let h be a function defined for all x≠0 such that h(4)=-3 and the derivative h is given by

h'(x)=(x^2-2)/(x) for all x≠0
a). Find all values of x for which the graph of h has a horizontal tangent, and determine whether h has a local maximum, a local minimum, or neither at each of these values. Justify your answers
b) On what intervals, if any, is the graph of h concave up? Justify your answers

## clearly the tangent is horizontal when

x^2-2 = 0

h"(x) = (x^2+2)/x^2

And you know that f(x) is concave up when f"(x) > 0

See that graph at

http://www.wolframalpha.com/input/?i=integral+x+-+2%2Fx

## a) To find the values of x for which the graph of h has a horizontal tangent, we need to find where the derivative h'(x) equals zero.

h'(x) = (x^2 - 2)/(x) = 0

We can ignore the denominator x since x ≠ 0. Solving the equation:

x^2 - 2 = 0

x^2 = 2

x = ±√2

Therefore, the graph of h has a horizontal tangent at x = √2 and x = -√2.

To determine whether h has a local maximum, local minimum, or neither at each of these values, we need to look at the behavior of the derivative h'(x) around these points.

For x < -√2, h'(x) is negative. For x between -√2 and √2, h'(x) is positive. And for x > √2, h'(x) is negative again.

From this information, we can conclude that at x = -√2, the graph of h has a local minimum, and at x = √2, the graph of h has a local maximum.

b) To find the intervals where the graph of h is concave up, we need to analyze the second derivative of h. Let's find h''(x):

h'(x) = (x^2 - 2)/(x)
h''(x) = [(2x)(x) - (x^2 - 2)]/(x^2)
h''(x) = (2x^2 - x^2 + 2)/(x^2)
h''(x) = (x^2 + 2)/(x^2)

The graph of h will be concave up when h''(x) > 0.

(x^2 + 2)/(x^2) > 0

Since x^2 is always positive except at x = 0 (which is not in the domain of h), we can ignore it and focus on the numerator (x^2 + 2).

x^2 + 2 > 0

This inequality is always true since the sum of a positive number (x^2) and a positive constant (2) is always positive.

Therefore, the graph of h is concave up for all values of x.

In summary:
a) The graph of h has horizontal tangents at x = √2 and x = -√2. At x = -√2, h has a local minimum, and at x = √2, h has a local maximum.
b) The graph of h is concave up for all values of x.

## To find the values of x for which the graph of h has a horizontal tangent, we need to find where the derivative of h is equal to zero.

Given that h'(x) = (x^2-2)/x, we set this equal to zero and solve for x:
(x^2-2)/x = 0

Since this expression can only be equal to zero if the numerator is equal to zero, we have:
x^2 - 2 = 0

Solving for x, we get:
x = ±√2

So, the graph of h has a horizontal tangent at x = √2 and x = -√2.

To determine whether h has a local maximum, local minimum, or neither at these values, we can use the first derivative test.

First, let's analyze the sign of h' to the left and right of x = √2 and x = -√2.

For x > √2:
Plugging in a value larger than √2 (e.g., x = 3), we have:
h'(3) = (3^2-2)/3 = (9-2)/3 > 0
This means that the derivative is positive on the right side of x = √2.

For x < √2:
Plugging in a value smaller than √2 (e.g., x = 1), we have:
h'(1) = (1^2-2)/1 = (1-2)/1 < 0
This means that the derivative is negative on the left side of x = √2.

Similarly, we can analyze the sign of h' to the left and right of x = -√2.

For x > -√2:
Plugging in a value larger than -√2 (e.g., x = -1), we have:
h'(-1) = (-1^2-2)/-1 = (1-2)/-1 > 0
This means that the derivative is positive on the right side of x = -√2.

For x < -√2:
Plugging in a value smaller than -√2 (e.g., x = -3), we have:
h'(-3) = (-3^2-2)/-3 = (9-2)/-3 < 0
This means that the derivative is negative on the left side of x = -√2.

Based on the sign of the derivative, we can conclude that:
- At x = √2, h has a local minimum.
- At x = -√2, h has a local maximum.

Now let's move on to part (b) of the question.

To determine where the graph of h is concave up, we need to analyze the sign of h''(x), the second derivative of h.

To find h''(x), we need to differentiate h'(x):
h''(x) = d/dx [(x^2-2)/x]
= [(2x*x - (x^2-2))/x^2]
= [2x^2 - x^2 + 2]/x^2
= [x^2 + 2]/x^2

For the graph of h to be concave up, h''(x) needs to be positive. Let's analyze the sign of h''(x) for different values of x:

- For x > 0, h''(x) = (x^2 + 2)/x^2 > 0
This means that the graph of h is concave up for x > 0.

- For x < 0, h''(x) = (x^2 + 2)/x^2 > 0
This means that the graph of h is concave up for x < 0.

So, the graph of h is concave up for all values of x, except for x = 0 (since the function h is not defined at x = 0).

In summary:
a) The graph of h has a horizontal tangent at x = √2 and x = -√2. At x = √2, h has a local minimum, and at x = -√2, h has a local maximum.
b) The graph of h is concave up for all values of x, except x = 0.

## To find the values of x where the graph of h has a horizontal tangent, we need to find where the derivative h'(x) is equal to zero or undefined.

First, let's set h'(x) = 0 and solve for x:

(x^2 - 2)/x = 0

To solve this equation, we multiply both sides by x:

x^2 - 2 = 0

Adding 2 to both sides:

x^2 = 2

Taking the square root of both sides:

x = ±√2

Therefore, the values of x where the derivative h'(x) is equal to zero are x = √2 and x = -√2.

Now, let's check whether h'(x) is undefined for any x values. The derivative h'(x) is undefined when the denominator x is equal to zero. However, we know from the given function that x ≠ 0, so we don't have to worry about undefined values.

To determine whether h has a local maximum, a local minimum, or neither at each of these values, we need to examine the behavior of h'(x) around those points.

Let's calculate h'(x) for some values that are close to √2 and -√2:

For x = √2 + 1:
h'(√2 + 1) = (√2 + 1)^2 - 2 / (√2 + 1)
= 2 + 2√2 + 1 - 2 / (√2 + 1)
= 3 + 2√2 / (√2 + 1)

Since both the numerator and denominator are positive, h'(√2 + 1) > 0, which means that the slope of the graph of h is positive when x is slightly greater than √2. Therefore, the graph of h has neither a local maximum nor a local minimum at x = √2.

Similarly, let's calculate h'(-√2 + 1):

h'(-√2 + 1) = (-√2 + 1)^2 - 2 / (-√2 + 1)
= 2 - 2√2 + 1 - 2 / (-√2 + 1)
= -3 - 2√2 / (-√2 + 1)

Since the numerator is negative and the denominator is positive, h'(-√2 + 1) < 0, which means that the slope of the graph of h is negative when x is slightly less than -√2. Therefore, the graph of h has neither a local maximum nor a local minimum at x = -√2.

In conclusion, the graph of h has a horizontal tangent at x = √2 and x = -√2, but it does not have a local maximum or a local minimum at these points.

Moving on to part b), to determine on what intervals the graph of h is concave up, we need to find where the second derivative h''(x) is positive.

To find h''(x), we take the derivative of h'(x):

h''(x) = d/dx [(x^2 - 2)/x]
= (2x * x - (x^2 - 2))/x^2
= (2x^2 - x^2 + 2)/x^2
= (x^2 + 2)/x^2

We want to find the values of x where h''(x) > 0.

(x^2 + 2)/x^2 > 0

The numerator x^2 + 2 is always positive, so the sign of h''(x) depends on the sign of x^2.

When x^2 > 0, h''(x) > 0.
When x^2 < 0, h''(x) < 0.

Since x ≠ 0, x^2 will always be positive or negative. Therefore, h''(x) is positive for all x except x = 0.

In conclusion, the graph of h is concave up for all x ≠ 0.