4

If $f(x) = \sin^4 x+\cos^2 x\;\forall x\; \in \mathbb{R}\;,$ Then $\bf{Max.}$ and $\bf{Min.}$ value of $f(x).$

My Solution:: Let $$\displaystyle y = \sin^4 x+\cos^2 x \leq \sin^2 x+\cos^2 x=1$$

And for Minimum, We take $$\displaystyle y = \sin^4 x+\cos^2 x=(1-\cos^2 x)^2+\cos^2 x$$

So $$\displaystyle y=\cos^4 x-\cos^2 x+1 = \left(\cos^2 x-\frac{1}{2}\right)^2+\frac{3}{4}\geq \frac{3}{4}$$

So We get $\displaystyle y=\sin^4 x+\cos^2 x\in \left[\frac{3}{4}\;,1\right]$

My question is How can we find Min. of $f(x)$ other then that method,

Something Like Using Inequality.,Thanks

Michael Hardy
  • 1
juantheron
  • 53,015
  • You meant $\left(\cos^2 x -\frac{1}{2}\right)^2$. Also, you have to show equality can be achieved for each of $1$ (when $x=0$) and $\frac{3}{4}$ (when $\cos^2 x=\frac{1}{2}$, which can be when $x=\frac{\pi}{6}$). Other than that, it is correct. – user26486 Apr 26 '15 at 13:54
  • @user31415 $\cos^2x=1/2$ is not satisfied by $x=\pi/6$; however it is satisfied by some $x$, because $-1\le 1/\sqrt{2}\le 1$. – egreg Apr 26 '15 at 15:38
  • @egreg It's satisfied by $x=\frac{\pi}{4}$, sorry. – user26486 Apr 26 '15 at 15:47

4 Answers4

5

We have \begin{align} f(x) & = \sin^4(x) + \cos^2(x) = \sin^4(x) + 1 - \sin^2(x) = 1 - \sin^2(x)(1-\sin^2(x))\\ & = 1-\sin^2(x)\cos^2(x) = 1 - \dfrac{\sin^2(2x)}4 \end{align} I trust you can finish off from here.

Adhvaitha
  • 20,259
2

$$\cos^4x-\cos^2x=\cos^2x(\cos^2x-1)=-\sin^2x\cos^2x=-\dfrac{(2\sin x\cos x)^2}4=-\dfrac{\sin^22x}4$$

For real $x,0\le\sin^22x\le1\iff-1\ge-\sin^22x\ge0$

In case you don't know double angle formula, $(\sin x\pm\cos x)^2\ge0\iff-1\le2\sin x\cos x\le1$

user26486
  • 11,331
lab bhattacharjee
  • 274,582
2

If we set $\sin^2 x=u\in[0,1]$ we have to find min/max of: $$ g(u) = u^4-u^2+1, \tag{1}$$ or just study the function: $$ f(v) = v^2-v+1 \tag{2}$$ over the interval $[0,1]$. So we just have a parabola with minimum attained in the vertex $v=\frac{1}{2}$ and maximum attained in the endpoints of $[0,1]$.

Jack D'Aurizio
  • 353,855
2

We can also solve this problem as follows.

Let $f(x)=\sin^4x+\cos^2x$. This simplifies $f(x)=\frac{1}{8}(7+\cos 4x).$ Then $f_{min}=\frac{3}{4}$ which occurs when $\cos 4x=-1$ and $f_{max}=1$ which occurs when $\cos 4x=1$.

Joseph Quarcoo
  • 324
  • 1
  • 5