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(a) If $ g(x) = x^6 + x^4 $ , $ x \ge 0 $, use a computer algebra system to find an expression for

$ g^{-1} (x) $.

(b) Use the expression in part (a) to graph $ y = g(x) $ , $ y = x $ , and $ y = g^{-1} (x) $ on the same screen.

a) $g^{-1}(x)=\sqrt{\frac{1}{3}\left[\sqrt[3]{\frac{3 \sqrt{3} \sqrt{27 x^{2}-4 x}+27 x-2}{2}}+\sqrt[3]{\frac{2}{3 \sqrt{3} \sqrt{27 x^{2}-4 x}+27 x-2}}\right]-\frac{1}{3}}$

b) $-\sqrt[3]{27 x+3 \sqrt{3} \sqrt{x(27 x-4)}-2}+\frac{2 \sqrt[3]{2}}{\sqrt[3]{27 x+3 \sqrt{3} \sqrt{x(27 x-4)}-2}}-2 \|$

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Oregon State University

Harvey Mudd College

University of Michigan - Ann Arbor

Idaho State University

Yeah, it's clear. So when you read here, so we have g of X B equal to X to the stick. Plus, next to the fourth, we substitute why we make this inverse. So we get X is equal to lie to the sick plus y to the fourth. We're going to use the caste system to sulfur. Why? And we actually get six solutions. We get three solutions, but all of them can be positive or negative, which makes a total of six solutions four of them we can't use since they contain an eye, which makes them non riel. And between the two, we can't use the negative one since X has to be positive or equal to zero. So we have the inverse be equal to square root of 1/3 Hugh Britton of three. We're rid of three square root of 27. Thanks. Square minus four X plus 27 X minus two over too. Plus cuba it, uh, two over three square root of three square of 27 X square, minus four cats last 27 X minus two two minus 1/3 for part B. We're going to draw a graph for organ autograph would look something like this. And then we have X is equal to why and then our inverse function.