Question

Let f(x) = x^3 + 2x + 6. In order to prove that f(x) ->10 as x -> 2,
complete the following steps.
(a) Show that f(x) - 10 = (x - 2)(x^2 + 2x + 2).
(b) If |x - 2| < 1, prove that |x^2 + 2x + 2|< 17

Answer 1

i already solved a)

Answer 2

ok we can do this

Answer 3

and this will complete the problem because you have control over |x-2| so you can make it less than epsilon over 17

Answer 4

tried squeeze theorem but no result

Answer 5

no need.
\[x^2+2x+2\] is a quadratic polynomial with vertex at x = -1, it is increasing on
\[(-1,\infty)\] and decreasing on
\[(-\infty,-1)\]

Answer 6

so if
\[|x-2|<1\] that means
\[-1<x-2<1\] or
\[1<x<3\]

Answer 7

since
\[x^2+2x+2\] is increasing on this interval, its minimum is at the left hand endpoint (at x = 1 you get 5) and maximum is at the right hand endpoint (at x = 3 you get 17)

Answer 8

clear?

Answer 9

ya man ill try it out thx

Answer 10

ok and if this is fancy calc class you can say "since on (1,3) the derivative is positive, we know the function is increasing on that interval so max is 17"
sounds fancier