The minimum value of 2^(x^2-3)^3+27 equals:

$(x^2-3)^3+27$

$f(x)=2$

$\log f(x)=(\log 2)((x^2-3)^3+27)$

differentiating wrt x

$\frac{1}{f(x)}f'(x)=3(x^2-2)^2×2x\log 2$

so $f'(x)=2^{(x^2-3)^3+27}×6(x^2-3)^2.x$

$f'(x)$ changes sign at $x=0$ from "-" to "+"

so min value is at x = 0

$f(0)=2^0=1$