Determine the concavity of
f(x)=xlnx
I end up with
F''(x)=1/x, where it would be concave up when x>0 and concave down when x<0
Is this correct?
y' = x(1/x) + lnx
= 1 + lnx
y'' = 1/x ---> yup, you had that
concave up when 1/x > 0 or x> 0
concave down when 1/x < 0, or x < 0
yes, you got it!
No, your reasoning is not correct. The function f(x) = xln(x) is only defined for x > 0, so there is no need to consider the concavity for x < 0.
To determine the concavity of f(x), you need to find the second derivative and analyze its sign:
1. Find the first derivative of f(x):
F'(x) = d/dx (xln(x))
= ln(x) + x * (1/x)
= ln(x) + 1
2. Find the second derivative of f(x):
F''(x) = d/dx (ln(x) + 1)
= 1/x
3. Now, we analyze the sign of F''(x):
- F''(x) > 0 for x > 0: This means that the function is concave up for x > 0.
- F''(x) = 0 when x = 0: However, as mentioned earlier, the function f(x) = xln(x) is not defined for x = 0.
- F''(x) is not defined for x < 0, as the original function is not defined for x < 0.
Therefore, f(x) = xln(x) is concave up for x > 0, and the concavity is not defined for x < 0.
To determine the concavity of a function, you need to analyze the second derivative of the function. In this case, let's start by finding the first and second derivatives of the function f(x) = xln(x).
First derivative (f'(x)):
To find f'(x), we can use the product rule. The product rule states that if you have a function f(x) = u(x)v(x), where u and v are both differentiable functions, the derivative is given by f'(x) = u'(x)v(x) + u(x)v'(x).
Using the product rule, u(x) = x and v(x) = ln(x), let's find the derivative:
f'(x) = (x)'ln(x) + x(ln(x))'
We know that the derivative of ln(x) is 1/x, so the equation becomes:
f'(x) = 1 * ln(x) + x(1/x)
Simplifying further, we get:
f'(x) = ln(x) + 1
Second derivative (f''(x)):
To find f''(x), we need to differentiate f'(x) with respect to x. Applying the derivative to f'(x) = ln(x) + 1:
f''(x) = (ln(x) + 1)'
Differentiating ln(x) gives us 1/x, and the derivative of a constant (such as 1) is zero, so we have:
f''(x) = 1/x + 0
Simplifying further:
f''(x) = 1/x
Now, let's analyze the concavity based on the sign of f''(x):
If f''(x) > 0, the function is concave up.
If f''(x) < 0, the function is concave down.
In our case, f''(x) = 1/x.
Since 1/x is positive for x > 0 and negative for x < 0, we can conclude that the function f(x) = xln(x) is concave up for x > 0 and concave down for x < 0.
Therefore, your conclusion is correct.