9.
f(x)= x sqrt(x^2+4) on interval [-5,4]
What region does it concave down?
What region does it concave up?
Max?
Min?
I don't think you will find many folks who will volunteer to do your work for you. Do you have any questions about the procedure, or method?
These questions seem pretty straightford exercises in use of the first and second derivatives, and the meaning thereof.
I agree with bob
You have posted 8 consecutive questions of a rather routine and elementary level.
I would consider these of mediocre difficulty that I would have expected most of my students to be able to do.
These are the type of question one would find as examples for this topic in most textbooks.
Show some effort and we will gladly help you with any difficulty you run into.
Oh well, go ahead, take away my calculus warmup for the morning :)
Hey mk
take the first derivative and set it to zero.
the resulting roots are critical points where the slope is horizontal
then take the second derivative at those points and look at the sign
if +, then that was a minimum
if negative, then that was a maximum
if zero, it flattens out, then resumes.
Thanks Damon.
And to Bob and Reiny, I was not asking for answers at all, I assumed that when someone looked at the question they would explain how to do it, not give me the answer. Considering I have all the answers in the back of my book. But thanks.
To determine the regions of concavity, maximum, and minimum of the function f(x) = x√(x^2+4), we need to find the second derivative of the function and analyze its sign.
Step 1: Find the first derivative
f'(x) = √(x^2+4) + (x^2+4)^(1/2)*(1/2)*(2x)
Step 2: Simplify the first derivative
f'(x) = √(x^2+4) + x*(x^2+4)^(-1/2)*(2x)
Step 3: Find the second derivative
f''(x) = [x*(x^2+4)^(-1/2)]*2 + 2*(x^2+4)^(1/2)*(2x)
= [2x*(x^2+4) + 4*(x^2+4)] / [(x^2+4)^(1/2)]
Step 4: Simplify the second derivative
f''(x) = [2x^3 + 8x + 16x] / [(x^2+4)^(1/2)]
= (2x^3 + 24x) / [(x^2+4)^(1/2)]
= (2x(x^2+12)) / [(x^2+4)^(1/2)]
Now we need to analyze the sign of f''(x) to determine the regions of concavity, maximum, and minimum.
Step 5: Determine the sign of f''(x)
We can consider three intervals: (-∞, -5), (-5, 4), and (4, +∞).
For the interval (-∞, -5):
Since x^2+12 is positive and (x^2+4)^(1/2) is always positive or zero, f''(x) = (2x(x^2+12)) / [(x^2+4)^(1/2)] is negative for all x in (-∞, -5).
Therefore, this is the region where f(x) is concave down.
For the interval (-5, 4):
Since x^2+12 is positive and (x^2+4)^(1/2) is always positive or zero, f''(x) = (2x(x^2+12)) / [(x^2+4)^(1/2)] is positive for all x in (-5, 4).
Therefore, this is the region where f(x) is concave up.
For the interval (4, +∞):
Since x^2+12 is positive and (x^2+4)^(1/2) is always positive or zero, f''(x) = (2x(x^2+12)) / [(x^2+4)^(1/2)] is negative for all x in (4, +∞).
Therefore, this is the region where f(x) is concave down.
Step 6: Identify the maximum and minimum
To find the maximum and minimum, we need to examine the critical points and endpoints of the interval [-5, 4].
Critical points occur where the first derivative is equal to zero or undefined. However, in our case, the first derivative is defined for all x, so it does not have any critical points. So we should check the endpoints of the interval instead.
For x = -5:
f(-5) = -5√(25+4) = -5√29
For x = 4:
f(4) = 4√(16+4) = 4√20 = 8√5
Hence, the maximum value of f(x) is 8√5, which occurs at x = 4, and the minimum value of f(x) is -5√29, which occurs at x = -5.
In summary:
Region of concave down: (-∞, -5) and (4, +∞)
Region of concave up: (-5, 4)
Maximum: f(4) = 8√5
Minimum: f(-5) = -5√29