anonymous
  • anonymous
Determine the values of x where the tangent line is horizontal for the function: f(x) = \frac{1}{(x^2 - 25)( x - 7)}. The value(s) of x where the tangent line to the graph of the function is horizontal is(are) .
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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ash2326
  • ash2326
Find f'(x) and equate it to 0, to find the value of x
amistre64
  • amistre64
latex only parses with the delimiters capping it
anonymous
  • anonymous
\[f(x) = \frac{1}{(x^2 - 25)( x - 7)}.\]

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More answers

ash2326
  • ash2326
for which tangent is horizontal or slope is zero
amistre64
  • amistre64
i almost thought of this as a max/min and was gonna say f'(x)=0 wasnt sufficient; but, as is it is correct to find horizontal tangents ;)
anonymous
  • anonymous
i know i have to find the derivative of f(x) which is the same as the tangent line
anonymous
  • anonymous
and then from the drivative solve for x
amistre64
  • amistre64
since a quotient rule still has to carry out the product rule; id just rewrite for product
anonymous
  • anonymous
but i tried this and didnt get the correct answer
anonymous
  • anonymous
what a pain. derivative is \[f'(x)=\frac{-3x^2+14x+25}{(x^2-25)^2(x-7)^2}\]so you have to set \[-3x^2+14x+25=0\] and use quadratic formula
anonymous
  • anonymous
so now i dont know what to do
anonymous
  • anonymous
so i got x = -1.37850958 and 6.04517624
amistre64
  • amistre64
ewww, id keep it in exact form
anonymous
  • anonymous
should i round these numbers?
anonymous
  • anonymous
ok....
anonymous
  • anonymous
so what do i do after this
amistre64
  • amistre64
you move on to the next problem. a good rule of thumb is: when you get to the end, stop.
anonymous
  • anonymous
but dont i have to find a y value too? how can i ?
amistre64
  • amistre64
"Determine the values of x where..."; so dont go creating any extra work for ourself unless you are really really excited
anonymous
  • anonymous
lol ok
anonymous
  • anonymous
thanks... i only needed the x values and math doesnt make me excited ...its just so difficult
amistre64
  • amistre64
;) good luck

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