Question

Let g(x)=x^3+ax+b, where a!=b. If the tangent lines to g at x=a and x=b are parallel, find g(1).

Answer 1

oh sorry

Answer 2

is it not that g(1) = 1+a +b?

Answer 3

sure, but I'm looking for a constant answer.

Answer 4

I don't see the link between the condition with the answer :(
g'(x) = 3x^2+a
so at x =a , g' (a) = 3a^2 +a
and g'(b) = 3b^2 +a
the tangent lines parallel shows 3a^2+a= 3b^2+a or a =\(\pm b\)
but it doesn't affect g(1) .
Ha!!

Answer 5

If it helps, the answer is 1.

Answer 6

if a = -b, then g(1) = 1
but we still have a =b,

Answer 7

@dumbcow

Answer 8

@Loser66 , you are correct and it is given that a does not equal b
so only option is that a = -b

Answer 9

oops, hehehe...

Answer 10

:D

Answer 11

Nice catch. Thanks, guys.

Answer 12

thank you. I forgot the condition.