find values a, b so h(x)= e^(ax) for x≤ 0, = √(x+b) for x>0, is differentiable at x=0

[fresh]

find values a, b so h(x)= e^(ax) for x≤ 0, = √(x+b) for x>0, is differentiable at x=0

h(x) = e^(ax) for x≤ 0
= √(x+b) for x>0

for which values of a and b will h(x) be differentiable at x= 0

if it is differentiable, d/dx [e^(ax)] should equal d/dx [√(x+b)]

d/dx [e^(ax)] = e^(ax) • a

d/dx [√(x+b)] = 1/ [2√(x+b)]


e^(ax) • a = 1/ [2√(x+b)]

at x = 0
e^(a0) • a = 1/ [2√(0+b)]
a = 1/2√b

is this the right set up for the problem? where do i go from here?

 

[tkhunny]

Two Objections:

1) It also has to be Continuous. You ignored this.
2) The square root does not exist at x = 0, so simply substituting is not acceptable.

One Suggestion:

1) Use both pieces of information and your confusions concerning where to go will disappear.

 

[fresh]

Two Objections:

1) It also has to be Continuous. You ignored this.
2) The square root does not exist at x = 0, so simply substituting is not acceptable.

One Suggestion:

1) Use both pieces of information and your confusions concerning where to go will disappear.

if the function is continuous e^(ax) =√(x+b) at x=0
e^(0) = √(b)
1 = √(b)
1= b


the slope from the left side must equal the slope from the right side at x= 0

e^(ax) • a = 1/ [2√(x+b)]


1 • a = 1/ [2√(b)]
a = 1/2√1 since b =1
a= 1/2

therefore for h(x) to be differentiable at x= 0
b=1 and a =1/2
is this right?

 

[tkhunny]

You tell me. Does it work? If so, you're done.