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Let $f(x)=a+(x-4)^{\frac{4}{9}}$, then minima of $f(x)$ is
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The correct answer is:
a
$\because f(x)=a+(x-4)^{4 / 9}$
$\therefore \quad f^{\prime}(x)=0+\frac{4}{9}(x-4)^{-5 / 9}$
Clearly, at $x=4, f^{\prime}(x)$ is not defined
Hence, $x=4$ is the point of extremum.
$\because \quad f(4)=a+(4-4)^{4 / 9}=a$
$\therefore$ The minimum value of $f(x)$ is $a$.
$\therefore \quad f^{\prime}(x)=0+\frac{4}{9}(x-4)^{-5 / 9}$
Clearly, at $x=4, f^{\prime}(x)$ is not defined
Hence, $x=4$ is the point of extremum.
$\because \quad f(4)=a+(4-4)^{4 / 9}=a$
$\therefore$ The minimum value of $f(x)$ is $a$.
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