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Let

$$f(x)=\left\{% \begin{array}{ll} x^3-x^2+10x-5,& x\leq 1 \\ -2x+\log_2 (b^2-2),& x>1 \\ \end{array}% \right.$$

Find all possible real values of $b$ such that $f(x)$ has the greatest value at $x=1$.

How do I proceed? I tried checking LHL=RHL, but I'm getting $b=\pm \sqrt{130}$. The answer is given as $b \in [-\sqrt{130},-2] \cup[\sqrt{2},\sqrt{130}]$.

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  • $\begingroup$ As the function is defined in two parts such that $x=1$ lies in the upper equation should you not use this one? $\endgroup$ – Henry Lee Aug 5 '18 at 14:53
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$$(x^3-x^2+10x-5)'=3x^2-2x+10>0$$ and it's obvious that $f$ decreases on $(1,+\infty).$

Thus, we need $$-2\cdot1+\log_2(b^2-2)\leq f(1)$$ or $$\log_2(b^2-2)\leq7$$ or $$0<b^2-2\leq128$$ or $$[-\sqrt{130},-\sqrt2)\cup(\sqrt2,\sqrt{130}].$$

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  • $\begingroup$ Thank you for the help $\endgroup$ – Arka Seth Aug 5 '18 at 15:42
  • $\begingroup$ You are welcome! $\endgroup$ – Michael Rozenberg Aug 5 '18 at 15:43
  • $\begingroup$ How does modifying the parameter $b$ change the value of $f(x)$ at $x=1$? The function $f$ does not depend on $b$ for $x \leq 1$. $\endgroup$ – Sambo Aug 5 '18 at 20:35
  • $\begingroup$ @Sambo For $x\leq1$ $f$ increases. See better my solution. $\endgroup$ – Michael Rozenberg Aug 5 '18 at 20:38
  • $\begingroup$ Oh, I understand my confusion. I understood the question as "what value of $b$ maximizes $f(1)$", as opposed to "what value of $b$ ensures that the global maximum of $f$ is at $x=1$". $\endgroup$ – Sambo Aug 5 '18 at 22:14

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