Given: #f(x) = { (e^(kx) " if " x >= 0), (kx+1 " if " x < 0) :}#. If #f(x)# is infinitely differentiable, then what is the value of #k# ?

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Answer 1 · 2018-02-21T12:55:29

#k = 0#

For #x < 0# we have:

#f''(x) = 0#

For #x >= 0# we have:

#f''(x) = k^2 e^(kx)#

So:

#lim_(x->0^-) f''(x) = 0#

#lim_(x->0^+) f''(x) = k^2 e^0 = k^2#

So we require #k^2 = 0# and thus #k=0#