Study monotonicity of non-rational function
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Let $y=g(frac{1}{ax})$, where $a>0$, $g=f^{-1}$, $f=F'$. I also know that $F:[0, infty) rightarrow [0,B] $ is strictly increasing, concave and bounded, with $F(0) =0$.
I want to study under what conditions $frac{x}{g(frac{1}{ax})}$ is increasing for $x>0$.
If I'm doing things right, taking derivatives I get: $$frac{g(frac{1}{ax})-xg'(frac{1}{ax})}{[g(frac{1}{ax})]^2},$$ where $g'(frac{1}{ax})=(1/f'(g(frac{1}{ax})))(-1/(ax^2)).$
Where do I go from here? Alternatively, can anybody suggest a better approach for this problem? Thanks!
calculus
asked Dec 4 at 16:20
User6548
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Let $y=g(frac{1}{ax})$, where $a>0$, $g=f^{-1}$, $f=F'$. I also know that $F:[0, infty) rightarrow [0,B] $ is strictly increasing, concave and bounded, with $F(0) =0$.
I want to study under what conditions $frac{x}{g(frac{1}{ax})}$ is increasing for $x>0$.
If I'm doing things right, taking derivatives I get: $$frac{g(frac{1}{ax})-xg'(frac{1}{ax})}{[g(frac{1}{ax})]^2},$$ where $g'(frac{1}{ax})=(1/f'(g(frac{1}{ax})))(-1/(ax^2)).$
Where do I go from here? Alternatively, can anybody suggest a better approach for this problem? Thanks!
calculus
asked Dec 4 at 16:20
User6548
11
Note that $text{dom } g$ is the image of $f$, which is decreasing, so $text{dom } g=(0,f'(0)]$. Then $0<1/axle f'(0)$ and $xge frac a{f'(0)}$.
– ajotatxe
Dec 4 at 16:34
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let $y=g(frac{1}{ax})$, where $a>0$, $g=f^{-1}$, $f=F'$. I also know that $F:[0, infty) rightarrow [0,B] $ is strictly increasing, concave and bounded, with $F(0) =0$.
I want to study under what conditions $frac{x}{g(frac{1}{ax})}$ is increasing for $x>0$.
If I'm doing things right, taking derivatives I get: $$frac{g(frac{1}{ax})-xg'(frac{1}{ax})}{[g(frac{1}{ax})]^2},$$ where $g'(frac{1}{ax})=(1/f'(g(frac{1}{ax})))(-1/(ax^2)).$
Where do I go from here? Alternatively, can anybody suggest a better approach for this problem? Thanks!
calculus
asked Dec 4 at 16:20
User6548
11
Let $y=g(frac{1}{ax})$, where $a>0$, $g=f^{-1}$, $f=F'$. I also know that $F:[0, infty) rightarrow [0,B] $ is strictly increasing, concave and bounded, with $F(0) =0$.
I want to study under what conditions $frac{x}{g(frac{1}{ax})}$ is increasing for $x>0$.
If I'm doing things right, taking derivatives I get: $$frac{g(frac{1}{ax})-xg'(frac{1}{ax})}{[g(frac{1}{ax})]^2},$$ where $g'(frac{1}{ax})=(1/f'(g(frac{1}{ax})))(-1/(ax^2)).$
Where do I go from here? Alternatively, can anybody suggest a better approach for this problem? Thanks!
calculus
calculus
asked Dec 4 at 16:20
User6548
11
asked Dec 4 at 16:20
User6548
11
asked Dec 4 at 16:20
User6548
11
asked Dec 4 at 16:20
User6548
11
asked Dec 4 at 16:20
User6548
11
11
Note that $text{dom } g$ is the image of $f$, which is decreasing, so $text{dom } g=(0,f'(0)]$. Then $0<1/axle f'(0)$ and $xge frac a{f'(0)}$.
– ajotatxe
Dec 4 at 16:34
add a comment |
Note that $text{dom } g$ is the image of $f$, which is decreasing, so $text{dom } g=(0,f'(0)]$. Then $0<1/axle f'(0)$ and $xge frac a{f'(0)}$.
– ajotatxe
Dec 4 at 16:34
Note that $text{dom } g$ is the image of $f$, which is decreasing, so $text{dom } g=(0,f'(0)]$. Then $0<1/axle f'(0)$ and $xge frac a{f'(0)}$.
– ajotatxe
Dec 4 at 16:34
Note that $text{dom } g$ is the image of $f$, which is decreasing, so $text{dom } g=(0,f'(0)]$. Then $0<1/axle f'(0)$ and $xge frac a{f'(0)}$.
– ajotatxe
Dec 4 at 16:34
add a comment |
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Note that $text{dom } g$ is the image of $f$, which is decreasing, so $text{dom } g=(0,f'(0)]$. Then $0<1/axle f'(0)$ and $xge frac a{f'(0)}$.
– ajotatxe
Dec 4 at 16:34