Needing help finding/setting up equation to find a horizontal tangent

irishpump

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Oct 25, 2011
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Thanks for any help:

Question:

Let g(x) be ln(f(x)) and f(x) a function with x-intercepts at x=2 and x=-4 and with horizontal tangents at x=3 and x=5. Additionally, g'(1)=1 and g(0)=1. Let g(x) be ln(f(x)). At what point(s) does g have a horizontal tangent?

I recognize I need to start by using the chain rule, but I'm really confused how to set this up. Is this how I would do it (or am I even close?): g(ln(2))=0? or g'(1(1/2)? My tiny brain is going to explode. Again, thank you for any help offered.
 
So points where g(x) have horizontal tangents are the points where g(x)=0\displaystyle g\prime(x)=0. Since g(x)=ln(f(x))\displaystyle g(x)=\ln(f(x)) then g(x)=f(x)f(x)\displaystyle g\prime(x)=\frac{f\prime(x)}{f(x)} using the chain rule.

So setting f(x)f(x)=0\displaystyle \frac{f\prime(x)}{f(x)}=0 means setting the numerator, or f(x)=0\displaystyle f\prime(x)=0. They then tell us that f(x) has horizontal tangents at x=3 and x=5 and thus this is where f(x)=0\displaystyle f\prime(x)=0. Therefore, the points where g(x) have horizontal tangents are x=3 and x=5.
 
Hello, irishpump!

Much of the given information is unnecessary . . .


f(x)\displaystyle f(x) is a function with x\displaystyle x-intercepts at x=2\displaystyle x=2 and x=4\displaystyle x=-4, .not needed
. . and with horizontal tangents at x=3\displaystyle x=3 and x=5.\displaystyle x=5.
Additionally, g(1)=1\displaystyle g'(1)=1 and g(0)=1.\displaystyle g(0)=1. .
not needed

Let g(x)=lnf(x)\displaystyle g(x) \:=\:\ln|f(x)|

At what point(s) does g(x)\displaystyle g(x) have a horizontal tangent?
First, we note that: .f(x)>0\displaystyle f(x) \,>\,0


There is a wealth of information given . . . We just have to dig it out.


f(x)\displaystyle f(x) has x\displaystyle x-intercepts at .x=2\displaystyle x = 2 and -4.\displaystyle \text{-}4.

. . This means: .{f(2)=0f(-4)=0}\displaystyle \begin{Bmatrix}f(2) &=& 0 \\ f(\text{-}4) &=& 0 \end{Bmatrix} .
not needed


f(x)\displaystyle f(x ) has horizontal tangents at x=3\displaystyle x = 3 and x=5.\displaystyle x = 5.

. . This means: .{f(3)=0f(5)=0}\displaystyle \begin{Bmatrix} f'(3) &=& 0 \\ f'(5) &=& 0 \end{Bmatrix} .[1]



Question: .When does g(x)=0?\displaystyle g'(x) \,=\,0\:?

We have: .g(x)=0f(x)f(x)=0f(x)=0\displaystyle g'(x) \,=\,0 \quad\Rightarrow\quad \dfrac{f'(x)}{f(x)} \,=\,0 \quad\Rightarrow\quad f'(x) \,=\,0


From [1], we have the answers: .x=3,  x=5\displaystyle x = 3,\;x = 5



Edit: srmichael beat me to it.
 
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