<s>[INST] Plotting graphs using numerical/mathematica method From the author's equation 13, 14 We can write by inserting V''(A)=0, 
Solving for R we get, 
$$R= \frac{6^{D/4} \sqrt{D}}{\sqrt{-2^{1+\frac{D}{2}} 3^{D/2}+3 2^{1+D} A-3^{1+\frac{D}{2}} A^2}}$$
Now inserting the  V into the article equation (11)$$E= \left(\frac{\pi }{2}\right)^{D/2} R^D V,$$ we get,
$$E= \left(\frac{\pi }{2}\right)^{D/2} \left(-\left(\frac{2}{3}\right)^{D/2} A^3+2^{\frac{1}{2} (-4-D)} A^4+A^2 \left(1+\frac{D}{2 R^2}\right)\right) R^D$$
Now inserting the value of R, we get, 
$$E= \left(-\left(\frac{2}{3}\right)^{D/2} A^3+2^{\frac{1}{2} (-4-D)} A^4+A^2 \left(1+2^{-1-\frac{D}{2}} 3^{-D/2} \left(-2^{1+\frac{D}{2}} 3^{D/2}+3 2^{1+D} A-3^{1+\frac{D}{2}} A^2\right)\right)\right) \left(\frac{6^{D/4} \sqrt{D}}{\sqrt{-2^{1+\frac{D}{2}} 3^{D/2}+3 2^{1+D} A-3^{1+\frac{D}{2}} A^2}}\right)^D \left(\frac{\pi }{2}\right)^{D/2}$$
For $D= 3$
 we finally get, 
$$E= \frac{27 6^{3/4} \left(-\frac{2}{3} \sqrt{\frac{2}{3}} A^3+\frac{A^4}{8 \sqrt{2}}+A^2 \left(1+\frac{-12 \sqrt{6}+48 A-9 \sqrt{3} A^2}{12 \sqrt{6}}\right)\right) \pi ^{3/2}}{\left(-12 \sqrt{6}+48 A-9 \sqrt{3} A^2\right)^{3/2}} \tag{1}$$
the graph for equation (1) must satisfy the article graph (FIG 2)
My graph:
Plot[(27 6^(3/4) (-(2/3) Sqrt[2/3] A^3 + A^4/(8 Sqrt[2]) +A^2 (1 + (-12 Sqrt[6] + 48 A - 9 Sqrt[3] A^2)/( 12 Sqrt[6]))) \[Pi]^(3/2))/(-12 Sqrt[6] + 48 A - 9 Sqrt[3] A^2)^(3/2), {A, 0.5, 2.5}]


But the author got, 
Output : 
Am I doing wrong in simulation? 
Then The author got like this in Fig 3

`
 [/INST] Look at your function in a much more simple way:
$$R(A) = \frac{\sqrt{12}}{\sqrt{24 A - 9 A^2-12}}$$
for $D=2$.  The function should have a horizontal tangent when the derivative of the radicand is zero:
$$\frac{d}{dA} (24 A - 9 A^2-12) = 0 \implies A = \frac{4}{3}$$
which seems to agree with your result.  Also, $R$ blows up when the denominator goes to zero, or when $A=2$ and $A=2/3$, which also agrees with your plot.  So far, I'd say your plot looks good.
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