#$&* course Mth 271 11/9 9p *********************************************
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Given Solution: `a The first function you evaluate is x^2 - 3x + 3. You then cube this result. So the breakdown to get f(g(x)) form is f(z) = z^3 g(x) = x^2 - 3x + 3. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):I think this is the same thing. I was following the format in the text. ------------------------------------------------ Self-critique Rating:2
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Given Solution: `a The first function you evaluate is x+1. You then take this result to the -5 power. So the breakdown to get f(g(x)) form is f(z) = z^-.5 g(x) = x+1. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I think this is the same - the g(x) matches. I just used u because that’s what the book used. ------------------------------------------------ Self-critique Rating:2 ********************************************* Question: `q 2b 7th edition 2.5.24 der of f(t) = (9t+2)^(2/3) by gen power rule YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: (6t+4/3)^(-⅓) confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a This function is of the form u^(2/3), where u = 9 t + 2. The derivative of f(t) = u^p is f ' (t) = p * u^(p-1) * u ', by the general power rule. Here p = 2/3, and u ' = (9t + 2)' = 9 so we have f ' (t) = p u^(p-1) * u ' = 2/3 * u^(2/3 - 1) * 9 = 6 * u^(-1/3), or since u = 9 t + 2 f ' (t) = 6 ( 9 t + 2)^(-1/3). ** STUDENT COMMENT: This explanation loses me. Is there any way you could explain it different. I am having a hard time following exactly where I messed up. INSTRUCTOR RESPONSE: The power function rule (x^n) ' = n x ^ (n - 1) could be written just as well using u as the variable; it would read (u^n) ' = n u ^ (n - 1), where the ' now means the derivative with respect to u. Thus (u^(2/3)) ' = 2/3 u^(2/3 - 1) = 2/3 u^(-1/3). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): It seems that if u = (9t+2) then you would plug (9t+2) into ⅔(9t+2)^(-⅓) which is what I did. I don’t understand why you multiplied by 6 instead of ⅔. ------------------------------------------------ Self-critique Rating:2
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Given Solution: `a Using inner function u = 25+x^2 and outer function = (u)^(-1/2) we get n u^(n-1) du/dx is -1/2 * u^(-3/2) * 2x = - x ( 25+x^2)^(-3/2). ** STUDENT QUESTION f(x) = (25 + x^2)^-1/2 f(x) = -1/2 (25 + x^2) (25 + x^2)' f(x) = (-12.5 - 1/2x^2) (2x) Is what I had for my answer. I still don't quite understand where I am messing up. Although, this section I am having trouble with a little bit. INSTRUCTOR RESPONSE you appear to occasionally overlook the power function rule in this case you have (u^(-1/2)) ' = -1/2 u^(-1/2 - 1) = -1/2 u^(-3/2) So your expression f(x) = -1/2 (25 + x^2) (25 + x^2)' should have read f(x) = -1/2 (25 + x^2)^(-3/2) * (25 + x^2)' That appears to be the only step you're missing, so it's clear you are on the right track. I'm confident you'll clear this up, but be sure to let me know if not. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): n u^(n-1) du/dx is -1/2 * u^(-3/2) * 2x = - x ( 25+x^2)^(-3/2). ** I don’t understand where the *2x comes from to the left of the = nor where the -x to the right of the = comes from. ------------------------------------------------ Self-critique Rating:2