The least value of 2^{(x^2 - 3)^3+27} is- | vedclass

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(C) Let $f(x) = 2^{(x^2 - 3)^3+27}$.To find the least value of $f(x)$,we need to find the least value of the exponent $g(x) = (x^2 - 3)^3 + 27$.Since

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$1$

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(C) Let $f(x) = 2^{(x^2 - 3)^3+27}$.To find the least value of $f(x)$,we need to find the least value of the exponent $g(x) = (x^2 - 3)^3 + 27$.Since the function $h(u) = u^3$ is a strictly increasing function,the range of $(x^2 - 3)^3$ depends on the range of $x^2 - 3$.We know that $x^2 \ge 0$ for all real $x$,so $x^2 - 3 \ge -3$.Let $u = x^2 - 3$. Then $u \in [-3, \infty)$.The function $g(u) = u^3 + 27$ is strictly increasing for $u \in [-3, \infty)$.The minimum value of $g(u)$ occurs at the smallest value of $u$,which is $u = -3$.Thus,$g_{min} = (-3)^3 + 27 = -27 + 27 = 0$.Therefore,the least value of $f(x) = 2^{g(x)}$ is $2^0 = 1$.

The least value of $2^{(x^2 - 3)^3+27}$ is-

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