Find any relative extrema and any points of inflection if they exist of f(x)=x^2+ ln x^2 showing calculus. Please show work in detail so I can follow. Thanks. The answer is no relative max or min and the points of inflection are (1,1) and (-1,1) but I'm having trouble getting there
f(x) = x^2 + lnx^2
f' = 2x + 2/x = 2(x^2+1)/x
f" = 2 - 2/x^2 = 2(x^2-1)/x^2
It's clear f' is never 0, so no min/max
f"=0 when x=1 or -1
so, the points of inflection are (-1,1) and (1,1)
Note that ln(x^2) is defined for all x not zero, since x^2>=0. So its domain includes -1, while 2ln(x) does not, even though we can usually simplify ln(x^2) = 2ln(x).
To find the relative extrema and points of inflection of the function f(x) = x^2 + ln(x^2), we need to first find its derivative.
Step 1: Find the derivative of f(x)
Let's differentiate f(x) with respect to x using the rules of differentiation.
f'(x) = (2x) + [(2/x)(x^2)]
= 2x + (2x/x)
= 2x + 2
Step 2: Find the critical points
To find the critical points, we set the derivative equal to zero and solve for x.
2x + 2 = 0
Subtracting 2 from both sides:
2x = -2
Dividing both sides by 2:
x = -1
So, the critical point is x = -1.
Step 3: Analyze the first derivative
To determine the intervals of increase or decrease, as well as the presence of relative extrema, we need to examine the sign of f'(x).
For x < -1, f'(x) > 0 since 2x + 2 > 0.
For -1 < x < ∞, f'(x) > 0 since 2x + 2 > 0.
This means that the function is increasing for all x values.
Step 4: Find the second derivative
To determine the points of inflection, we need to find the second derivative of f(x).
f''(x) = d/dx(2x + 2)
= 2
Step 5: Set the second derivative equal to zero
To find the possible points of inflection, we set the second derivative equal to zero and solve for x.
2 = 0
Since 2 is never equal to zero, there are no points of inflection.
Step 6: Analyze the second derivative
Since the second derivative, f''(x) = 2, is always positive, it means that the function f(x) = x^2 + ln(x^2) is concave up for all x values.
Therefore, the conclusion is:
- There are no relative extrema (maxima or minima).
- There are no points of inflection.
It seems like the previous answer you mentioned (no relative max or min, and points of inflection at (1,1) and (-1,1)) is incorrect based on the provided function and the calculus analysis performed above.