Question

Find the critical numbers of the function g(x) = 8x^{2/3}(x^2 - 4). (Use symbolic notation and fractions where needed. Give your answer in the form of a comma-separated list of numbers. Enter NO SOLUTION if there are no solutions.) the critical number(s): Find all the local maximum and minimum points of g(x) = 8x^{2/3}(x^2 - 4). (Use symbolic notation and fractions where needed. Give your answers in the form of a comma-separated lists of points: (, ), (, ), ..., (, ). Enter NO SOLUTION if there are no solutions.) local maximum point(s): local minimum point(s):

          Find the critical numbers of the function g(x) = 8x^{2/3}(x^2 - 4).
(Use symbolic notation and fractions where needed. Give your answer in the form of a comma-separated list of numbers. Enter NO SOLUTION if there are no solutions.)
the critical number(s):
Find all the local maximum and minimum points of g(x) = 8x^{2/3}(x^2 - 4).
(Use symbolic notation and fractions where needed. Give your answers in the form of a comma-separated lists of points: (*, *), (*, *), ..., (*, *). Enter NO SOLUTION if there are no solutions.)
local maximum point(s):
local minimum point(s):

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$Find the critical numbers of the function g(x) = 8x^2/3(x^2 - 4). (Use symbolic notation and fractions where needed. Give your answer in the form of a comma-separated list of numbers. Enter NO SOLUTION if there are no solutions.) the critical number(s): Find all the local maximum and minimum points of g(x) = 8x^2/3(x^2 - 4). (Use symbolic notation and fractions where needed. Give your answers in the form of a comma-separated lists of points: (*, *), (*, *), ..., (*, *). Enter NO SOLUTION if there are no solutions.) local maximum point(s): local minimum point(s):$

Added by George F.

Calculus: Early Transcendentals

James Stewart 8th Edition

Instant Answer

Solved by Expert Gregory Higby

09/19/2023

Step 1

g'(x) = d/dx [8x^3(x^2 - 4)] Using the product rule, we get: g'(x) = 8x^3 d/dx (x^2 - 4) + (x^2 - 4) d/dx (8x^3) g'(x) = 8x^3(2x) + (x^2 - 4)(24x^2) g'(x) = 16x^4 + 24x^2(x^2 - 4) Now, we need to find the critical numbers by setting g'(x) = 0: 16x^4 + 24x^2(x^2 Show more…

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Find the critical numbers of the function g(x) = 8x^{2/3}(x^2 - 4). (Use symbolic notation and fractions where needed. Give your answer in the form of a comma-separated list of numbers. Enter NO SOLUTION if there are no solutions.) the critical number(s): Find all the local maximum and minimum points of g(x) = 8x^{2/3}(x^2 - 4). (Use symbolic notation and fractions where needed. Give your answers in the form of a comma-separated lists of points: (*, *), (*, *), ..., (*, *). Enter NO SOLUTION if there are no solutions.) local maximum point(s): local minimum point(s):

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00:01 So what we have is the function defined as 8x to the 2 3rds power times x squared minus 4.

00:10 And what i would do is start with the derivative of this.

00:13 And you have a product going on so you have to do the product rule well.

00:16 The derivative of the left side gives me 16 3rds x to the negative 1 3rd power.

00:23 You leave the right side alone, x squared minus 4.

00:26 And then plus, now you take the derivative of the right side, which would be 2x.

00:31 And you leave the left side alone of 8x to the 2 3rds power.

00:37 I think what i would do is try to just get this so it's a little bit cleaned up.

00:45 That you have x to the 1 3rd power here.

00:47 You also have a 3, 16, x squared minus 4.

00:52 But then here you would even have 16x.

00:57 Adding the exponents would be the 5 3rds power.

01:00 But if you're trying to get the same denominator, so you can add them together, you might put 3x to the 1 3rd here, which means you also have to multiply the top by 3x to the 1 3rd power.

01:13 And now you can add those exponents.

01:15 So if i've done my math correctly, let's go ahead and distribute here.

01:20 16x squared minus 64.

01:24 And then this would be 16 times 3 is 48.

01:27 Adding those exponents is 6 3rds, which is x squared.

01:32 What i want to see is where that equals 0.

01:34 And i can add 64 to the right side.

01:37 I believe 16 plus 48 is 64x squared.

01:42 And so x squared could equal 1 if i divide over...

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