Question

Calculus help, I am trying to do this questions.. someone teach me please 10 points

Answer 1

Explanation:

Answer 2

A bit late, but better than never.

24. For
`x\\eq1`, we have

`(9x^2 - x - 8)/(x - 1) = ((x-1)(9x+8))/(x-1) = 9x+8`

Then as
`x` approaches 1, we have

`\displaystyle \lim_(x\to1) f(x) = \lim_(x\to1) (9x+8) = 17 \\eq 0`

so the function is not continuous at
`x=1`. It is thus continuous on the interval union
`(-\infty,0)\cup(0,\infty)`. You can also write this as "
`x<0` or
`x>0`".

25. When
`x` approaches 2 from the left (when
`x<2`), we have

`\displaystyle \lim_(x\to2^-) f(x) = \lim_(x\to2) (5-x) = 3`

When
`x` approaches 2 from the right
`(x>2)`, we have

`\displaystyle \lim_(x\to2^+) f(x) = \lim_(x\to2) (2x-3) = 1`

so the function is not continuous at
`x=2`. Thus it's continuous on
`(-\infty,2)\cup(2,\infty)`, or "
`x<2` or
`x>2`".