Least gradient value for $f(x)=e^{-2x} tan x$
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I’ve derived it into $e^{-2x} (-1+ tan x)^2$, but i couldn’t think of any other way other than to look at a graph which is π/4. Is there any way to find it through calculation?
calculus algebra-precalculus derivatives trigonometry
edited Dec 14 '18 at 11:20
gimusi
92.9k84494
asked Dec 14 '18 at 11:18
WangcincayWangcincay
463
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add a comment |
$begingroup$
I’ve derived it into $e^{-2x} (-1+ tan x)^2$, but i couldn’t think of any other way other than to look at a graph which is π/4. Is there any way to find it through calculation?
calculus algebra-precalculus derivatives trigonometry
edited Dec 14 '18 at 11:20
gimusi
92.9k84494
asked Dec 14 '18 at 11:18
WangcincayWangcincay
463
$endgroup$
add a comment |
$begingroup$
I’ve derived it into $e^{-2x} (-1+ tan x)^2$, but i couldn’t think of any other way other than to look at a graph which is π/4. Is there any way to find it through calculation?
calculus algebra-precalculus derivatives trigonometry
edited Dec 14 '18 at 11:20
gimusi
92.9k84494
asked Dec 14 '18 at 11:18
WangcincayWangcincay
463
$endgroup$
I’ve derived it into $e^{-2x} (-1+ tan x)^2$, but i couldn’t think of any other way other than to look at a graph which is π/4. Is there any way to find it through calculation?
calculus algebra-precalculus derivatives trigonometry
calculus algebra-precalculus derivatives trigonometry
edited Dec 14 '18 at 11:20
gimusi
92.9k84494
asked Dec 14 '18 at 11:18
WangcincayWangcincay
463
edited Dec 14 '18 at 11:20
gimusi
92.9k84494
asked Dec 14 '18 at 11:18
WangcincayWangcincay
463
edited Dec 14 '18 at 11:20
gimusi
92.9k84494
edited Dec 14 '18 at 11:20
gimusi
92.9k84494
edited Dec 14 '18 at 11:20
gimusi
92.9k84494
92.9k84494
asked Dec 14 '18 at 11:18
WangcincayWangcincay
463
asked Dec 14 '18 at 11:18
WangcincayWangcincay
463
asked Dec 14 '18 at 11:18
WangcincayWangcincay
463
463
add a comment |
add a comment |
2 Answers
2
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Yes your derivation is correct indeed we have
$$f(x)=e^{-2x} tan x implies f'(x)=-2e^{-2x} tan x+e^{-2x} (1+tan^2 x)=$$$$=e^{-2x}left(tan^2 x-2tan x+1right)=e^{-2x}left(tan x-1right)^2ge 0$$
where we have used that
- $e^{-2x}>0$
- $left(tan x-1right)^2ge 0$
Can you figure out when $f'(x)=0$?
answered Dec 14 '18 at 11:26
gimusigimusi
92.9k84494
$endgroup$
add a comment |
$begingroup$
You correctly differentiated to get $f’(x) = e^{-2x}(tan x-1)^2$. Notice how both factors are non-negative:
$$e^{-2x} > 0$$
$$(tan x-1)^2 geq 0$$
Hence, the least value must be $geq 0$.
By inspection, it’s clear how the second factor can become $0$:
$$tan x-1 = 0$$
from which you obtain the desired result.
answered Dec 14 '18 at 11:28
KM101KM101
6,0601525
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add a comment |
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2 Answers
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2 Answers
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active
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$begingroup$
Yes your derivation is correct indeed we have
$$f(x)=e^{-2x} tan x implies f'(x)=-2e^{-2x} tan x+e^{-2x} (1+tan^2 x)=$$$$=e^{-2x}left(tan^2 x-2tan x+1right)=e^{-2x}left(tan x-1right)^2ge 0$$
where we have used that
- $e^{-2x}>0$
- $left(tan x-1right)^2ge 0$
Can you figure out when $f'(x)=0$?
answered Dec 14 '18 at 11:26
gimusigimusi
92.9k84494
$endgroup$
add a comment |
$begingroup$
Yes your derivation is correct indeed we have
$$f(x)=e^{-2x} tan x implies f'(x)=-2e^{-2x} tan x+e^{-2x} (1+tan^2 x)=$$$$=e^{-2x}left(tan^2 x-2tan x+1right)=e^{-2x}left(tan x-1right)^2ge 0$$
where we have used that
- $e^{-2x}>0$
- $left(tan x-1right)^2ge 0$
Can you figure out when $f'(x)=0$?
answered Dec 14 '18 at 11:26
gimusigimusi
92.9k84494
$endgroup$
add a comment |
$begingroup$
Yes your derivation is correct indeed we have
$$f(x)=e^{-2x} tan x implies f'(x)=-2e^{-2x} tan x+e^{-2x} (1+tan^2 x)=$$$$=e^{-2x}left(tan^2 x-2tan x+1right)=e^{-2x}left(tan x-1right)^2ge 0$$
where we have used that
- $e^{-2x}>0$
- $left(tan x-1right)^2ge 0$
Can you figure out when $f'(x)=0$?
answered Dec 14 '18 at 11:26
gimusigimusi
92.9k84494
$endgroup$
Yes your derivation is correct indeed we have
$$f(x)=e^{-2x} tan x implies f'(x)=-2e^{-2x} tan x+e^{-2x} (1+tan^2 x)=$$$$=e^{-2x}left(tan^2 x-2tan x+1right)=e^{-2x}left(tan x-1right)^2ge 0$$
where we have used that
- $e^{-2x}>0$
- $left(tan x-1right)^2ge 0$
Can you figure out when $f'(x)=0$?
answered Dec 14 '18 at 11:26
gimusigimusi
92.9k84494
answered Dec 14 '18 at 11:26
gimusigimusi
92.9k84494
answered Dec 14 '18 at 11:26
gimusigimusi
92.9k84494
answered Dec 14 '18 at 11:26
gimusigimusi
92.9k84494
92.9k84494
add a comment |
add a comment |
$begingroup$
You correctly differentiated to get $f’(x) = e^{-2x}(tan x-1)^2$. Notice how both factors are non-negative:
$$e^{-2x} > 0$$
$$(tan x-1)^2 geq 0$$
Hence, the least value must be $geq 0$.
By inspection, it’s clear how the second factor can become $0$:
$$tan x-1 = 0$$
from which you obtain the desired result.
answered Dec 14 '18 at 11:28
KM101KM101
6,0601525
$endgroup$
add a comment |
$begingroup$
You correctly differentiated to get $f’(x) = e^{-2x}(tan x-1)^2$. Notice how both factors are non-negative:
$$e^{-2x} > 0$$
$$(tan x-1)^2 geq 0$$
Hence, the least value must be $geq 0$.
By inspection, it’s clear how the second factor can become $0$:
$$tan x-1 = 0$$
from which you obtain the desired result.
answered Dec 14 '18 at 11:28
KM101KM101
6,0601525
$endgroup$
add a comment |
$begingroup$
You correctly differentiated to get $f’(x) = e^{-2x}(tan x-1)^2$. Notice how both factors are non-negative:
$$e^{-2x} > 0$$
$$(tan x-1)^2 geq 0$$
Hence, the least value must be $geq 0$.
By inspection, it’s clear how the second factor can become $0$:
$$tan x-1 = 0$$
from which you obtain the desired result.
answered Dec 14 '18 at 11:28
KM101KM101
6,0601525
$endgroup$
You correctly differentiated to get $f’(x) = e^{-2x}(tan x-1)^2$. Notice how both factors are non-negative:
$$e^{-2x} > 0$$
$$(tan x-1)^2 geq 0$$
Hence, the least value must be $geq 0$.
By inspection, it’s clear how the second factor can become $0$:
$$tan x-1 = 0$$
from which you obtain the desired result.
answered Dec 14 '18 at 11:28
KM101KM101
6,0601525
edited Dec 14 '18 at 11:37
answered Dec 14 '18 at 11:28
KM101KM101
6,0601525
answered Dec 14 '18 at 11:28
KM101KM101
6,0601525
answered Dec 14 '18 at 11:28
KM101KM101
6,0601525
6,0601525
add a comment |
add a comment |
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