Question $(1)$minimum value of $\displaystyle f(x) = \frac{\sqrt{25+x^2}}{2}+\frac{6-x}{4}$

Question $(2)$ A function $f: R \rightarrow R$ is defined as $f(x) = |x|^m\cdot |x-1|^n\;\forall x \in [0,1] \in R$ and $m,n \in N$

then maximum value of function

i am trying to find out maximum and minimum value of above questions without using derivative but not be able to evaluate , could some help me

  • 1
    $\begingroup$ ALL CAPITAL nicks looks bad, as if you would continuously shouting. I suggest a change to simply "Durgesh Tiwari". $\endgroup$ – peterh Feb 12 '17 at 8:14
  • 2
    $\begingroup$ For $(1),$ set $x=5\tan y$ $\endgroup$ – lab bhattacharjee Feb 12 '17 at 8:22
  • 1
    $\begingroup$ For (2) there is no maximum on $\mathbb{R}$: the expression can be arbitrarily large ! You should restrict to values $0\leq x \leq1$... $\endgroup$ – Jean Marie Feb 12 '17 at 8:33

For (1) let $g(x)=2(\sqrt {x^2+25})-x=4f(x)-6.$

(i). We have $g(x)\geq 2(\sqrt {x^2})-x=2|x|-x\geq 0. $

(ii). $g(x)=y \iff$ $ y+x=2\sqrt {x^2+25}\iff [y+x\geq 0\;\land \; (y+x)^2=4x^2+100]\iff$ $$(\bullet)\quad [y+x\geq 0\;\land \; 0=(100-y^2)-2xy+3x^2.]$$

The quadratic equation $0=(100-y^2)-2xy+3x^2$ has a real solution in $x$ iff the discriminant $$4y^2-12(100-y^2)\geq 0,$$ that is, iff $|y|\geq 5\sqrt 3,$ iff $y\geq 5\sqrt 3 $ (because $y=g(x)$ is never negative).

Therefore $f(x)=(6+g(x))/4\geq (6+5\sqrt 3\;)/4.$

This minimum value is achieved when $y=5\sqrt 3$. Putting $y=5\sqrt 3$ in $(\bullet)$ gives $(3x-5\sqrt 3\;)^2/3=0$. So $(5/3)\sqrt 3$ is the value of $x$ that minimizes $f(x)$.


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.