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       Can someone tell me how to do this? 

 Oct 27, 2016
 #1
avatar+257 
0

Uhhh idk man really i diont that is to advanced for rn ok

angelangel

angelangel

 Oct 27, 2016
 #2
avatar+129850 
0

Not sure what you want to do......here's the graph : https://www.desmos.com/calculator/0lpspyn1ud

 

 

cool cool cool

 Oct 27, 2016
 #3
avatar+1904 
0

\(f(x)=\sqrt[3]{{x}^{2}+4x}\)

 

\({f(x)}^{3}={\sqrt[3]{{x}^{2}+4x}}^{3}\)

 

\({f(x)}^{3}={x}^{2}+4x\)

 

\({f(x)}^{3}=x(x+4)\)

 

\(\frac{{f(x)}^{3}}{x}=\frac{x(x+4)}{x}\)

 

\(\frac{{f(x)}^{3}}{x}=x+4\)

 

\(\frac{{f(x)}^{3}}{x}-4=x+4-4\)

 

\(\frac{{f(x)}^{3}}{x}-4=x+0\)

 

\(\frac{{f(x)}^{3}}{x}-4=x\)

 

\(x(\frac{{f(x)}^{3}}{x}-4)=x\times x\)

 

\({f(x)}^{3}-4x=x\times x\)

 

\({f(x)}^{3}-4x={x}^{2}\)

 

\({f(x)}^{3}-4x-{x}^{2}={x}^{2}-{x}^{2}\)

 

\({f(x)}^{3}-4x-{x}^{2}={0x}^{2}\)

 

\({f(x)}^{3}-4x-{x}^{2}=0\)

 

\(-{x}^{2}-4x+{f(x)}^{3}=0\)

 

\(x = {4 \pm \sqrt{(-4)^2-4(-1)({f(x)}^{3})} \over 2(-1)}\)

 

\(x = {4 \pm \sqrt{16-4(-1)({f(x)}^{3})} \over 2(-1)}\)

 

\(x = {4 \pm \sqrt{16-(-4)({f(x)}^{3})} \over 2(-1)}\)

 

\(x = {4 \pm \sqrt{16+4{f(x)}^{3}} \over 2(-1)}\)

 

\(x = {4 \pm \sqrt{4(4+{f(x)}^{3})} \over 2(-1)}\)

 

\(x = {4 \pm 2\sqrt{4+{f(x)}^{3}} \over 2(-1)}\)

 

\(x = {2(2 \pm 1\sqrt{4+{f(x)}^{3}} )\over 2(-1)}\)

 

\(x = {2(2 \pm \sqrt{4+{f(x)}^{3}} )\over 2(-1)}\)

 

\(x = {2(2 \pm \sqrt{4+{f(x)}^{3}} )\over -2}\)

 

\(x = -1(2 \pm \sqrt{4+{f(x)}^{3}} )\)

 

\(x = -(2 \pm \sqrt{4+{f(x)}^{3}} )\)

 

\(x = -2 \pm \sqrt{4+{f(x)}^{3}}\)

.
 Oct 27, 2016

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