- #1
Jeff Ford
- 155
- 2
I am working to prove that this function is continuous at [itex] x = 2 [/itex]
[tex] {f(x) = 9x - 7} [/tex]
To do this I know that I have to show that [itex] \vert f(x) – f(a) \vert < \epsilon [/itex] and that [itex] \vert x - a < \delta \vert [/itex]
I tried to come up with a relationship between [itex] \vert x - 2 \vert [/itex] and [itex] \epsilon [/itex] so I could get an appropriate number to choose for [itex] \delta [/itex]
This is as far as I got
[tex] \vert {f(x) – f(a)} \vert < \epsilon [/tex]
[tex] \vert {9x – 7} \vert < \epsilon [/tex]
I’m stuck. All of the examples the text shows give equations where it is easy to factor out the [itex] \vert {x - a} \vert [/itex] term.
A push in the right direction would be appreciated.
[tex] {f(x) = 9x - 7} [/tex]
To do this I know that I have to show that [itex] \vert f(x) – f(a) \vert < \epsilon [/itex] and that [itex] \vert x - a < \delta \vert [/itex]
I tried to come up with a relationship between [itex] \vert x - 2 \vert [/itex] and [itex] \epsilon [/itex] so I could get an appropriate number to choose for [itex] \delta [/itex]
This is as far as I got
[tex] \vert {f(x) – f(a)} \vert < \epsilon [/tex]
[tex] \vert {9x – 7} \vert < \epsilon [/tex]
I’m stuck. All of the examples the text shows give equations where it is easy to factor out the [itex] \vert {x - a} \vert [/itex] term.
A push in the right direction would be appreciated.
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