For what value(s)?

For what value(s) of p and q is the function:
enter image source here
differentiable?

1 Answer
Andrea S.
Mar 1, 2018

#p=5#

#q=1#

Explanation:

The function is continuous and differentiable for # x < 1# and for #x >1# and:

#f'(x) = {(3x^2+2 " if " x > 1) , (p " if " x < 1):}#

For the function to be continuous and differentiable in #x =1# the value we get for #f(x)# and #f'(x)# when #x->1# from the left and from the right must be the same.

So:

#lim_(x->1^-) f(x) = lim_(x->1^+) f(x)#

#lim_(x->1^-) px+q = lim_(x->1^+) x^3+2x+3#

#p+q = 6#

which is the condition making #f(x)# continuous, and:

#lim_(x->1^-) f'(x) = lim_(x->1^+) f'(x)#

#p= lim_(x->1^+) 3x^2+2#

#p=5#

which is the condition making #f'(x)# continuous, and therefore #f(x)# differentiable also for #x=1#

In conclusion:

#{(p=5),(q=1):}#

Related questions
Impact of this question
340 views around the world
You can reuse this answer
Creative Commons License