1. ## 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

2. 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 $\displaystyle \leq 1$ or $\displaystyle \geq 1$ in the function ?

3. Actually it is f(x) = e^(ax), if x is less than or equal to one.
Thanks.

4. The problem is to find $\displaystyle a$ because for $\displaystyle b$ is $\displaystyle b= e^{a}$. The equation for $\displaystyle a$ is...

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

... that can be written as...

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

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

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

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

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$

5. 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.

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

Thanks

6. Thank you so much!