If $f(x)=3 x+\frac{12}{x}$ is continuous on $\mathbb{R}-\{0\}$ and $M$ is its maximum value, then $\lim _{x \rightarrow M}(f x)=$

Question

Aicharya · Accepted Answer

$\because f(x)=3 x+\frac{12}{x} \Rightarrow f^{\prime}(x)=3-\frac{12}{x^2}$ For critical point: $f^{\prime}(x)=0$ $\begin{aligned} & \Rightarrow 3-\frac{12}{x^2}=0 \Rightarrow x^2=4 \Rightarrow x=-2,2 \ & f^{\prime \prime}(x)=\frac{24}{x^3}\end{aligned}$ At $x=-2, f^{\prime \prime}(-2)=-3 < 0$ $	herefore f(x)$ has maxima at $x=-2$ and maximum value is $\begin{aligned} & f(-2)=-6-6=-12 \ & 	herefore M=-12\end{aligned}$ $\lim _{x ightarrow M} f(x)=\lim _{x ightarrow-12}\left(3 x+\frac{12}{x}ight)=-37$

If $f(x)=3 x+\frac{12}{x}$ is continuous on $\mathbb{R}-\{0\}$ and $M$ is its maximum value, then $\lim _{x…

Solution