- #1

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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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