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

[irishpump]

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.

 

[srmichael]

So points where g(x) have horizontal tangents are the points where \(\displaystyle g\prime(x)=0\). Since \(\displaystyle g(x)=\ln(f(x))\) then \(\displaystyle g\prime(x)=\frac{f\prime(x)}{f(x)}\) using the chain rule.

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

 

[soroban]

Hello, irishpump!

Much of the given information is unnecessary . . .

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

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

At what point(s) does \(\displaystyle g(x)\) have a horizontal tangent?

First, we note that: .\(\displaystyle f(x) \,>\,0\)


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


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

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


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

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



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

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


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


Edit: srmichael beat me to it.