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sketch a graph with given properties of $f'(x)$
$begingroup$
Sketch a graph with given properties:
continuous everywhere except for where a vertical asymptote at $x=0$,
$f'(x)<0$ for $-infty <x<0$, $f'(x)>0$ for $0<x<infty$, $f''(x)<0$ for $-infty<x<0$
calculus
asked Oct 17 '15 at 20:24
Carrie StreettCarrie Streett
1
$endgroup$
add a comment |
$begingroup$
Sketch a graph with given properties:
continuous everywhere except for where a vertical asymptote at $x=0$,
$f'(x)<0$ for $-infty <x<0$, $f'(x)>0$ for $0<x<infty$, $f''(x)<0$ for $-infty<x<0$
calculus
asked Oct 17 '15 at 20:24
Carrie StreettCarrie Streett
1
$endgroup$
add a comment |
$begingroup$
Sketch a graph with given properties:
continuous everywhere except for where a vertical asymptote at $x=0$,
$f'(x)<0$ for $-infty <x<0$, $f'(x)>0$ for $0<x<infty$, $f''(x)<0$ for $-infty<x<0$
calculus
asked Oct 17 '15 at 20:24
Carrie StreettCarrie Streett
1
$endgroup$
Sketch a graph with given properties:
continuous everywhere except for where a vertical asymptote at $x=0$,
$f'(x)<0$ for $-infty <x<0$, $f'(x)>0$ for $0<x<infty$, $f''(x)<0$ for $-infty<x<0$
calculus
calculus
asked Oct 17 '15 at 20:24
Carrie StreettCarrie Streett
1
asked Oct 17 '15 at 20:24
Carrie StreettCarrie Streett
1
edited Oct 18 '15 at 0:41
user99914
asked Oct 17 '15 at 20:24
Carrie StreettCarrie Streett
1
asked Oct 17 '15 at 20:24
Carrie StreettCarrie Streett
1
asked Oct 17 '15 at 20:24
Carrie StreettCarrie Streett
1
1
add a comment |
add a comment |
1 Answer
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$begingroup$
Since $f'(x)$ is negative in the interval $(-infty,0)$, you know that $f(x)$ is decreasing in this interval. Likewise, since $f'(x)$ is positive in the interval $(0,infty)$, $f(x)$ needs to increase.
Since $f(x)$ has a vertical asymptote at $x=o$, the value of $f(x)$ increases approaching $|infty$ or $-infty$ as $x$ approaches $0$. This is represented by a sharp increase or decrease near $0$.
Because $f''(x)<0$ in the interval $(-infty,0)$, $f(x)$ should be decreasing at an increasing rate.
Using these hints should help you create a graph.
answered Oct 17 '15 at 20:57
JedJed
719414
$endgroup$
add a comment |
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1 Answer
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1 Answer
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$begingroup$
Since $f'(x)$ is negative in the interval $(-infty,0)$, you know that $f(x)$ is decreasing in this interval. Likewise, since $f'(x)$ is positive in the interval $(0,infty)$, $f(x)$ needs to increase.
Since $f(x)$ has a vertical asymptote at $x=o$, the value of $f(x)$ increases approaching $|infty$ or $-infty$ as $x$ approaches $0$. This is represented by a sharp increase or decrease near $0$.
Because $f''(x)<0$ in the interval $(-infty,0)$, $f(x)$ should be decreasing at an increasing rate.
Using these hints should help you create a graph.
answered Oct 17 '15 at 20:57
JedJed
719414
$endgroup$
add a comment |
$begingroup$
Since $f'(x)$ is negative in the interval $(-infty,0)$, you know that $f(x)$ is decreasing in this interval. Likewise, since $f'(x)$ is positive in the interval $(0,infty)$, $f(x)$ needs to increase.
Since $f(x)$ has a vertical asymptote at $x=o$, the value of $f(x)$ increases approaching $|infty$ or $-infty$ as $x$ approaches $0$. This is represented by a sharp increase or decrease near $0$.
Because $f''(x)<0$ in the interval $(-infty,0)$, $f(x)$ should be decreasing at an increasing rate.
Using these hints should help you create a graph.
answered Oct 17 '15 at 20:57
JedJed
719414
$endgroup$
add a comment |
$begingroup$
Since $f'(x)$ is negative in the interval $(-infty,0)$, you know that $f(x)$ is decreasing in this interval. Likewise, since $f'(x)$ is positive in the interval $(0,infty)$, $f(x)$ needs to increase.
Since $f(x)$ has a vertical asymptote at $x=o$, the value of $f(x)$ increases approaching $|infty$ or $-infty$ as $x$ approaches $0$. This is represented by a sharp increase or decrease near $0$.
Because $f''(x)<0$ in the interval $(-infty,0)$, $f(x)$ should be decreasing at an increasing rate.
Using these hints should help you create a graph.
answered Oct 17 '15 at 20:57
JedJed
719414
$endgroup$
Since $f'(x)$ is negative in the interval $(-infty,0)$, you know that $f(x)$ is decreasing in this interval. Likewise, since $f'(x)$ is positive in the interval $(0,infty)$, $f(x)$ needs to increase.
Since $f(x)$ has a vertical asymptote at $x=o$, the value of $f(x)$ increases approaching $|infty$ or $-infty$ as $x$ approaches $0$. This is represented by a sharp increase or decrease near $0$.
Because $f''(x)<0$ in the interval $(-infty,0)$, $f(x)$ should be decreasing at an increasing rate.
Using these hints should help you create a graph.
answered Oct 17 '15 at 20:57
JedJed
719414
edited Oct 18 '15 at 0:43
user99914
answered Oct 17 '15 at 20:57
JedJed
719414
answered Oct 17 '15 at 20:57
JedJed
719414
answered Oct 17 '15 at 20:57
JedJed
719414
719414
add a comment |
add a comment |
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