The derivative of f(x)=(x^4/3)-(x^5/5) attains its maximum value at what x value?
Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga.
Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus.
Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
Not the answer you are looking for? Search for more explanations.
f''(x) = 4x^2 - 3x^3
Isn't that it
Ummm, looks like your \(f''(x)\) is a bit off.
The condition To attain Max is
f''(x) > 0
That is true
They want the max of the derivative.
So you want to find the critical numbers \(f''(x)=0\) or undefined. Then you'd plug them back into \(f'(x)\) to see where the max is.
nope @wio they ask..For where the max is attained
"The derivative of f(x)=(x^4/3)-(x^5/5) attains its maximum"
"f'(x) attains its maximum"
aha! f"(x) = 4x^2 - 4x^3.
i got my critical numbers. x = 0 & x = 1?
That looks good. So your local maximum value is at either x=0 or x=1. Also, since your first derivative is a negative quartic (has a \(-x^4\) term), you have a parabola type shape opening downwards, so the local maximum is the global maximum. Now you just need to determine which one is the maximum.
make sense so far?
i think. i got f'(0) = 0 and f'(1) = 1/3. so the answer is... 1?
All riiiiiiiiight. thank you. and wio. but s/he left.