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

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 $b \in [-\sqrt{130},-2] \cup[\sqrt{2},\sqrt{130}]$.

  • $\begingroup$ 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

$$(x^3-x^2+10x-5)'=3x^2-2x+10>0$$ 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 $$[-\sqrt{130},-\sqrt2)\cup(\sqrt2,\sqrt{130}].$$

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.