Question

What is the maximum and min value of the function m(x)=3x^4−4x^3−12x^2 on the interval [−23] ?

Answer 1

interval -2,3 not -23

Answer 2

use extreme value theorem

Answer 3

continuous functions on a compact set attain an absolute max and min

Answer 4

The derivative of that function has zeroes at -1, 0, and 2, and the second derivative is negative at zero. So checking the endpoints and the local maximum, m(-2) = 32, m(0) = 0, and m(3) = 27. Therefore, your maximum value on that interval would be m(-2) = 32. Similarly, the second derivative is negative at -1 and 2, and m(-1) = -5 and m(2) = -32, so your minimum value would be at m(2) = -32.

Answer 5

Oops, I meant "Similarly, the second derivative is positive at -1 and 2......"

Answer 6

lets see

take the first derivative to get m'(x)=12x^3-12x^2-24x
setting m'(x) to zero,
12x(x^2-1x-2)=0
x = 0 for which y = 0
or x^2-x-2=0

x = (1+ (1+8)^(1/2))/2 =2 or (1-(1+8)^(1/2) / 2 =-1
substitute for value of x to get y. which gives you the critical points. Evaluate the function's value at these points for a comparison.

Answer 7

you don't need the 2nd derivative

Answer 8

find the critical numbers that lie within the set you are interested in and then evaluate the original function at those critical numbers and the endpoints of your interval. pick the smallest and largest

Answer 9

i have a whole bunch of calculus that i need help with

Answer 10

m(x)=3x^4−4x^3−12x^2

for x =2 y = 48-32-48 = -32
for x = -1 y =3-4+12=11

Answer 11

Given that this appears to be a first-semester calculus question I don't think it's wholly unreasonable to mention that the second derivative can distinguish between minima and maxima. The fact that this particular question is asking for both does mean that you need to evaluate all the points, but in general that's not necessarily true if they ask only for a min or a max and it's reasonably good practice to know whether to expect a min or a max when you're just starting out, I think.

Answer 12

What is the maximum and min value of f(x)=t2/t2+3 on the interval [−1,1] ?

Answer 13

I assume that's
\[f(x) = \frac{t^2}{t^2 + 3} \]
right?

Answer 14

yes

Answer 15

Well,
\[f'(x) = \frac{(t^2+3)(2t) - t^2(2t)}{(t^2+3)^2} = \frac{6t}{(t^2+3)^2} \]
Which is zero only at t=0. From that you can conclude that f(x) has either a local minimum or a local maximum at that point. Evaluating at the endpoints, as well as at that point,
\[f(-1) = 1/4, f(1) = 1/4, f(0) = 0\]
so it appears that the function attains its maximum value (1/4) at the endpoints and its minimum value (0) at t=0.

Answer 16

do you have time for a few more ?

Answer 17

Ask them as separate questions, not as follow ups to this thread, so other people have the opportunity to answer you.

Answer 18

usually the extreme value theorem appears before the 1st and 2nd derivative tests so talking about using the 1st or 2nd D-test to find relative extrema could be confusing.

Answer 19

The extreme value theorem is the basis for those tests, but in and of itself it doesn't tell you where you get a minimum or a maximum, only that you will, so I'm not sure what you mean when you say to use it to find those values.

Answer 20

In the context of real analysis and the theoretical foundations of calculus its an important theorem but it's intuitively obvious to a toddler with a piece of spaghetti.

Answer 21

The EVT is obviously needed here (along with Fermat's theorem), but the 2nd derivative is not needed.