# Calculus/Differentiability

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• February 18th 2010, 07:04 AM
pantera
Calculus/Differentiability
Hello, I need help finishing this problem. Use one-sided limits to find values of the constants a and b that make the piecewise function differentiable at the point where the function rule changes.

f(x) = e^(ax), if x <1
and
f(x) = b + ln x, if x>1

Here's what I have so far.....

For f to be continuous at x = 1,
lim as x goes to 1 from the left of e^(ax)= lim as x goes to 1 from the right of (b+ ln x)

implies e^(a)=b

For f to be differentiable at x = 1,
lim as x goes to 1 from the left of ae^(ax)=lim as x goes to 1 from the right of (1/x)

implies ae^(a) = 1

So I'm having trouble solving for a and b for these two equations....
e^(a)=b
ae^(a) = 1

The solution says solve by grapher: a = 0.5671....and b = 1.7632....,
but I don't know how to get that.

Thanks
• February 18th 2010, 07:15 AM
Ted
Quote:

Originally Posted by pantera
Hello, I need help finishing this problem. Use one-sided limits to find values of the constants a and b that make the piecewise function differentiable at the point where the function rule changes.

f(x) = e^(ax), if x <1
and
f(x) = b + ln x, if x>1

Here's what I have so far.....

For f to be continuous at x = 1,
lim as x goes to 1 from the left of e^(ax)= lim as x goes to 1 from the right of (b+ ln x)

implies e^(a)=b

For f to be differentiable at x = 1,
lim as x goes to 1 from the left of ae^(ax)=lim as x goes to 1 from the right of (1/x)

implies ae^(a) = 1

So I'm having trouble solving for a and b for these two equations....
e^(a)=b
ae^(a) = 1

The solution says solve by grapher: a = 0.5671....and b = 1.7632....,
but I don't know how to get that.

Thanks

You function is not defined at 1.
Are you sure there is no $\leq 1$ or $\geq 1$ in the function ?
• February 18th 2010, 07:20 AM
pantera
Actually it is f(x) = e^(ax), if x is less than or equal to one.
Thanks.
• February 18th 2010, 11:28 PM
chisigma
The problem is to find $a$ because for $b$ is $b= e^{a}$. The equation for $a$ is...

$a\cdot e^{a}=1$ (1)

... that can be written as...

$f(a)=a+\ln a=0$ (2)

An efficient way to solve an equation like (2) is the Newton-Raphson method that computes the sequence...

$a_{n+1} = a_{n} - \frac{f(a_{n})}{f^{'}(a_{n})}$ (3)

If $a_{0}$ is chosen correctly [for example $a_{0}=\frac {1}{2}$...] the sequence (3) converges to the solution...

Kind regards

$\chi$ $\sigma$
• February 19th 2010, 03:22 AM
HallsofIvy
Quote:

Originally Posted by pantera
Hello, I need help finishing this problem. Use one-sided limits to find values of the constants a and b that make the piecewise function differentiable at the point where the function rule changes.

f(x) = e^(ax), if x <1
and
f(x) = b + ln x, if x>1

Here's what I have so far.....

For f to be continuous at x = 1,
lim as x goes to 1 from the left of e^(ax)= lim as x goes to 1 from the right of (b+ ln x)

implies e^(a)=b

For f to be differentiable at x = 1,
lim as x goes to 1 from the left of ae^(ax)=lim as x goes to 1 from the right of (1/x)

implies ae^(a) = 1

So I'm having trouble solving for a and b for these two equations....
e^(a)=b
ae^(a) = 1

There is no "algebraic" way to solve that second equation. You can say that a= W(1) where W is the "Lambert W function" which is defined as the inverse to f(x)= xe^x. Other that that, it must be solved numerically which you seem to have done.

Quote:

The solution says solve by grapher: a = 0.5671....and b = 1.7632....,
but I don't know how to get that.

Thanks
• February 19th 2010, 07:08 AM
pantera
Thank you so much!