-3
$\begingroup$

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
    ALL CAPITAL nicks looks bad, as if you would continuously shouting. I suggest a change to simply "Durgesh Tiwari". – 2017-02-12
  • 2
    For $(1),$ set $x=5\tan y$ – 2017-02-12
  • 1
    For (2) there is no maximum on $\mathbb{R}$: the expression can be arbitrarily large ! You should restrict to values $0\leq x \leq1$... – 2017-02-12

1 Answers 1

1

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 $f(x)$.