#$&* course Mth 271 Question: `q 1b 7th edition 2.5.2 inner, outer fns for (x^2-3x+3)^3
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Given Solution: `a The first function you evaluate is x^2 - 3x + 3. You then cube this result. So the breakdown to get f(g(x)) form is f(z) = z^3 g(x) = x^2 - 3x + 3. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Self-critique Rating: ********************************************* Question: `q 1e 7th edition 2.5.8 inner, outer fns for (x+1)^-.5 YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Inner function is g(x) = x+1 Outer function is [g(x)]^-.5 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q 2b 7th edition 2.5.24 der of f(t) = (9t+2)^(2/3) by gen power rule YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 2/3(9t+2) ^-1/3 * 9 = [18(9t + 2)^-1/3]/3 = 18/3cuberoot(9t +2) confidence Rating:3
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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). ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): *** I assume that it is OK to write the solution the way I did it. I am not sure what is correct way to type 3rd root…. But I hope the idea is right. ------------------------------------------------ Self-critique Rating:
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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). ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: `q 2c 7th edition 2.5.28 der of f(x) = (25+x^2)^(-1/2) by gen power rule YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: -1/2 (25 + x^2) ^ -3/2 * (2x) = [-2x (25 + x^2) ^ -3/2]/2 = = -x (25 + x^2) ^ -3/2 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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). ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!