find the interval on which the graph of f(x)=ln(x^2+1) is concave upward.
f'=1/(x^2+1) *2x
f"=2/(x^2+1)- 4x^2/(x^2+1)^2 check that.
now, when is f" zero?
f"=0=> 2=4x^2/(x^2+1)
2x^2=x^2+1
x=+-1 So pick three points, say x=-2, x=0, and x=2 and see what f" is, is it positive?
Check my work, it is easy to make an error on a keyboard.
To find the interval on which the graph of f(x) = ln(x^2+1) is concave upward, we need to consider the second derivative of the function.
First, let's find the first derivative of f(x):
f'(x) = d/dx [ln(x^2+1)]
Using the chain rule, we have:
f'(x) = 1/(x^2+1) * d/dx(x^2+1)
= 1/(x^2+1) * (2x)
Next, let's find the second derivative of f(x):
f''(x) = d/dx[f'(x)]
= d/dx[1/(x^2+1) * (2x)]
Using the product rule, we have:
f''(x) = [1/(x^2+1)*d/dx(2x)] + [(2x)*d/dx(1/(x^2+1))]
Simplifying, we get:
f''(x) = (2/(x^2+1)) + [(2x)*(-2x)/(x^2+1)^2]
= (2/(x^2+1)) - (4x^2/(x^2+1)^2)
Now, we need to find the values of x for which f''(x) > 0 in order to determine the interval on which the graph of f(x) is concave upward.
Setting f''(x) > 0, we have:
(2/(x^2+1)) - (4x^2/(x^2+1)^2) > 0
Multiplying every term by (x^2+1)^2, we get:
2(x^2+1) - 4x^2 > 0
Expanding and simplifying, we have:
2x^2 + 2 - 4x^2 > 0
-2x^2 + 2 > 0
Dividing every term by -2, we have:
x^2 - 1 < 0
Factoring, we get:
(x + 1)(x - 1) < 0
Now, we can determine the sign of the expression (x + 1)(x - 1) for different intervals:
-∞ < x < -1: (x + 1) < 0, (x - 1) < 0, so (x + 1)(x - 1) > 0
-1 < x < 1: (x + 1) > 0, (x - 1) < 0, so (x + 1)(x - 1) < 0
1 < x < ∞: (x + 1) > 0, (x - 1) > 0, so (x + 1)(x - 1) > 0
Therefore, the interval on which the graph of f(x) = ln(x^2+1) is concave upward is -1 < x < 1.
To find the interval on which the graph of the function f(x) = ln(x^2 + 1) is concave upward, we need to examine the second derivative of the function. The second derivative will indicate the concavity of the function.
First, let's find the derivative of f(x):
f'(x) = d/dx[ln(x^2+1)]
To do this, we can use the chain rule:
f'(x) = (1 / (x^2 + 1)) * (2x)
Simplifying this, we get:
f'(x) = 2x / (x^2 + 1)
Now, let's find the second derivative of f(x):
f''(x) = d/dx [f'(x)]
Using the quotient rule, the second derivative is:
f''(x) = [(2 * (x^2 + 1)) - (2x * 2x)] / (x^2 + 1)^2
Simplifying further, we get:
f''(x) = (2 * (x^2 + 1) - 4x^2) / (x^2 + 1)^2
f''(x) = (2 - 2x^2) / (x^2 + 1)^2
Now, for the concavity to be upward, we need f''(x) > 0.
Setting the second derivative greater than zero:
2 - 2x^2 > 0
Rearranging the inequality:
2x^2 - 2 < 0
x^2 - 1 < 0
Factoring the quadratic expression:
(x - 1)(x + 1) < 0
The inequality holds true when:
-1 < x < 1
Thus, the graph of f(x) = ln(x^2 + 1) is concave upward in the interval (-1, 1).