For the function below, determine the intervals of concavity:
f(x)=xlnx
I tried doing it but the second derivative would always be f''(x)=1/x, so it would be concave up at x>0 and concave down at x<0, which is wrong.
Also, for the function below, determine the intervals of increase:
f(x)=2^(x/3)
I can't seem to isolate the x to find the critical points?
Find all the critical points and determine whether each is a local maximum or minimum or neither:
f'(x)=x(9-x^2)^(1/2)
I was just wondering if the maxes are at -3 and 3?
while f" = 1/x, f is not defined for x<=0. So, f is concave up for all x>0.
f = 2^(x/2)
f' = ln2/2 2^(x/2)
f" = (ln2/2)^2 2^(x/2)
This is just e^kx in disguise. It is concave up everywhere. There are no critical points.
f' = x√(9-x^2)
f'=0 at x=0, x=±3
f" = (9-2x^2)/√(9-x^2)
Since f">0 at x=0, that is a minimum
Since f" is undefined at x=±3, neither min nor max.
play around with these functions at wolframalpha.com to see the graphs.
1st question:
The answer is just that
"The graph is concave upward when x > 0"
Your approach to find the second derivative is correct and your second derivative is correct.
However, you should note that when x<0,
ln(x) is undefined, i.e. the original function is undefined when x <0 and so as the second derivative. (Domain of the function: x>0
That means you should not consider the concavity for x < 0 and that is the wrong thing about your answer.
To determine the intervals of concavity for a function, you need to analyze the second derivative. However, in the case of the function f(x) = xlnx, the second derivative f''(x) = 1/x is always positive for x > 0 and always negative for x < 0. This means that f(x) is concave up for all x > 0 and concave down for all x < 0.
Now, onto the second question regarding the function f(x) = 2^(x/3). To determine the intervals of increase, you need to find the derivative of the function and then solve for x when the derivative is greater than zero. However, finding the critical points in this case can be a bit challenging since the variable x is in the exponent.
To isolate x and find critical points for f(x) = 2^(x/3), you can take the logarithm of both sides. Applying the logarithm property, you get ln(f(x)) = ln(2^(x/3)).
Next, use the logarithm property that states ln(a^b) = b * ln(a) to simplify the equation: ln(f(x)) = (x/3) * ln(2).
Now, you have successfully isolated the x term. To find the critical points, differentiate ln(f(x)) with respect to x to get:
f'(x)/f(x) = (1/3) * ln(2).
Simplifying further, you'll arrive at f'(x) = (1/3) * f(x) * ln(2).
At this point, you can equate f'(x) to zero and solve for x to find any critical points.