Smallest Value

[NeedAssistance30]

I'm having trouble with the algebra and trig in the problem. I'm confused about substitution.

Find the smallest value of the function f(x)=3secx+4cscx on the interval (0,pi/2)

solve for df/dx=0

df/dx=3 tanxsecx-4cotxcscx= 3sinx/cos^2x-4cosx/sin^2x

 

[soroban]

Hello, NeedAssistance30!

$\displaystyle \text{Find the smallest value of the function: }f(x) \;=\; 3\sec x+4\csc x\:\text{ on the interval }\,(0,\,\tfrac{\pi}{2})$

$\displaystyle \text{Solve: }\:\frac{df}{dx} \:=\:0$

. . $\displaystyle \frac{df}{dx} \:=\:3\tan x\sec x-4\cot x\csc x\:=\: \frac{3\sin x}{\cos^2\!x} - \frac{4\cos x}{\sin^2\!x} \:=\:0$

. . . All correct!

$\displaystyle \text{We have: }\:\frac{3\sin x}{\cos^2\!x} \:=\:\frac{4\cos x}{\sin^2\!x} \quad\Rightarrow\quad \frac{\sin^3\!x}{\cos^3\!x} \:=\:\frac{4}{3} \quad\Rightarrow\quad \left(\frac{\sin x}{\cos x}\right)^3 \:=\:\frac{4}{3}$


. . . . . . . $\displaystyle \tan^3\!x \:=\:\tfrac{4}{3} \quad\Rightarrow\quad \tan x \:=\:\sqrt[3]{\tfrac{4}{3}} \quad\Rightarrow\quad x \:=\:\tan^{-1}\left(\sqrt[3]{\tfrac{4}{3}}\right)$

. $\displaystyle \text{Hence: }\:x \;\approx\;8.333$


$\displaystyle \text{Therefore: }\;f(0.8333) \;=\;3\sec(0.8333) + 4\csc(0.8333) \;=\;9.865$

 

[needmathhelp99]

How do you get to the sin^3x/cos^3x=3/4

How do you get to the [FONT=MathJax_Main]sin[/FONT][FONT=MathJax_Main]3[/FONT][FONT=MathJax_Math]x[/FONT][FONT=MathJax_Main]cos[/FONT][FONT=MathJax_Main]3[/FONT][FONT=MathJax_Math]x[/FONT]= 3/4
thanks

 

[needmathhelp99]

aren't you not supposed to cross mulitple?

 

[pka]

How do you get to the [FONT=MathJax_Main]sin[/FONT][FONT=MathJax_Main]3[/FONT][FONT=MathJax_Math]x[/FONT][FONT=MathJax_Main]cos[/FONT][FONT=MathJax_Main]3[/FONT][FONT=MathJax_Math]x[/FONT]= 3/4

Do not tack a new question onto an existing thread. Do you understand?

$\displaystyle \dfrac{{{{\sin }^3}(x)}}{{{{\cos }^3}(x)}} = {\tan ^3}(x)$

Therefore, $\displaystyle x =\displaystyle \arctan \left( {\sqrt[3]{{\frac{3}{4}}}} \right)$

 

[needmathhelp99]

Yes, but how do you go from the [FONT=MathJax_Main]3[/FONT][FONT=MathJax_Main]sin[/FONT][FONT=MathJax_Math]x [/FONT][FONT=MathJax_Main]cos[/FONT][FONT=MathJax_Main]2[/FONT][FONT=MathJax_Math]x[/FONT][FONT=MathJax_Main]= [/FONT][FONT=MathJax_Main]4cos[FONT=MathJax_Math]x/si^[/FONT][FONT=MathJax_Math]2x to sin^3/cos^3=4/3[/FONT]sin2[FONT=MathJax_Math]x/[/FONT][/FONT][FONT=MathJax_Main] sin[/FONT][FONT=MathJax_Main]2[/FONT][FONT=MathJax_Math]x[/FONT]

 

[pka]

Yes, but how do you go from the [FONT=MathJax_Main]3[/FONT][FONT=MathJax_Main]sin[/FONT][FONT=MathJax_Math]x [/FONT][FONT=MathJax_Main]cos[/FONT][FONT=MathJax_Main]2[/FONT][FONT=MathJax_Math]x[/FONT][FONT=MathJax_Main]= [/FONT][FONT=MathJax_Main]4cos[FONT=MathJax_Math]x/si^[/FONT][FONT=MathJax_Math]2x to sin^3/cos^3=4/3[/FONT]sin2[FONT=MathJax_Math]x/[/FONT][/FONT][FONT=MathJax_Main] sin[/FONT][FONT=MathJax_Main]2[/FONT][FONT=MathJax_Math]x[/FONT]

You did not post that that question. Please re-read what you actually posted.

Start a new thread! Post the question that you actually mean to ask.

Thank you.

 

[needmathhelp99]

I'm sorry; I didn't mean to be rude, and I apologize if I was an any way.
I do not understand how to make a new thread because I can't find the +Post New thread button as the FAQ says there is. I must not see it. I must have mistyped typed but I wanted to know how to get to the

[FONT=MathJax_Main]sin[/FONT][FONT=MathJax_Main]3[/FONT][FONT=MathJax_Math]x[/FONT][FONT=MathJax_Main]cos[/FONT][FONT=MathJax_Main]3[/FONT][FONT=MathJax_Math]x[/FONT][FONT=MathJax_Main]=[/FONT][FONT=MathJax_Main]4/[/FONT][FONT=MathJax_Main]3[/FONT]


I'm sorry,
Please forgive me.

 

[JeffM]

I'm sorry; I didn't mean to be rude, and I apologize if I was an any way.
I do not understand how to make a new thread because I can't find the +Post New thread button as the FAQ says there is. I must not see it. I must have mistyped typed but I wanted to know how to get to the

[FONT=MathJax_Main]sin[/FONT][FONT=MathJax_Main]3[/FONT][FONT=MathJax_Math]x[/FONT][FONT=MathJax_Main]cos[/FONT][FONT=MathJax_Main]3[/FONT][FONT=MathJax_Math]x[/FONT][FONT=MathJax_Main]=[/FONT][FONT=MathJax_Main]4/[/FONT][FONT=MathJax_Main]3[/FONT]


I'm sorry,
Please forgive me.

When you come into a sub-forum, there will be a button toward the the top left of your screen that says POST NEW THREAD

 

[needmathhelp99]

Thank you.