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A)

Find the value of k such that \( f(x) = \left\{\begin{array}{ll} x^2 + 1 &\text{if } x \le 1, \\ kx & \text{if } x > 1. \end{array}\right. \)

is differentiable at x=1.

 

B)

Find a function f such that f is not differentiable at x=0, but \(f^2\) is differentiable at x=0. (Note that \(f^2(x)= (f(x))^2\).)

 

Can someone please help me with these questions?

Thanks!

 Oct 30, 2020
 #1
avatar+9142 
+2

A)

 

Let   f1(x)  =  x2 + 1   and let   f2(x)  =  kx

 

In order for  f(x)  to be differentiable at  x = 1, two things must be true at the same time:

 

First, the graphs must "touch" at  x = 1.  That is,   f1(1)  =  f2(1)

 

Second, the slopes of both graphs at the point where they touch must be identical.  That is,   f1'(1)  =  f2'(1)

 

We can start with either condition and find which value of  k  makes it true, and then check to make sure that same value of  k  makes the other condition true. Let's start with the first condition:

 

f1(1)  =  f2(1)

                            And     f1(1)  =  12 + 1     and     f2(1)  =  k(1)

12 + 1  =  k(1)

                            Simplify both sides of the equation.

2  =  k

 

So we know that  k = 2  satisfies the first condition. Now let's check to make sure it also satisfies the second condition:

 

f1'(1)  =  f2'(1)

                            f1'(x)  =  2x     so     f1'(1)  =  2(1)  =  2     and     f2'(x)  =  k  =  2

2  =  2

 

The second condition is also satisfied when  k = 2.

 

So     k = 2

 

Check:  https://www.desmos.com/calculator/bitwqu3gbj

 Oct 31, 2020
 #2
avatar+9142 
+2

B)

 

f(x)  =  √x

 

Then  f  is not differentiable at  x = 0  because  f'(x)  =  \(\frac12x^{-\frac12}=\frac{1}{2\sqrt{x}}\)   which is undefined when  x = 0

 

And

 

f2(x)  =  ( √x )2   =   x

 

Which is differentiable at  x = 0  because  f2'(x)  =  1  for all values of  x .

 Oct 31, 2020

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