Question

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) .

Answer 1

Find f'(x) and equate it to 0, to find the value of x

Answer 2

latex only parses with the delimiters capping it

Answer 3

\[f(x) = \frac{1}{(x^2 - 25)( x - 7)}.\]

Answer 4

for which tangent is horizontal or slope is zero

Answer 5

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 ;)

Answer 6

i know i have to find the derivative of f(x) which is the same as the tangent line

Answer 7

and then from the drivative solve for x

Answer 8

since a quotient rule still has to carry out the product rule; id just rewrite for product

Answer 9

but i tried this and didnt get the correct answer

Answer 10

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

Answer 11

so now i dont know what to do

Answer 12

so i got x = -1.37850958 and 6.04517624

Answer 13

ewww, id keep it in exact form

Answer 14

should i round these numbers?

Answer 15

ok....

Answer 16

so what do i do after this

Answer 17

you move on to the next problem. a good rule of thumb is: when you get to the end, stop.

Answer 18

but dont i have to find a y value too? how can i ?

Answer 19

"Determine the values of x where..."; so dont go creating any extra work for ourself unless you are really really excited

Answer 20

lol ok

Answer 21

thanks... i only needed the x values and math doesnt make me excited ...its just so difficult

Answer 22

;) good luck