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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 $$f^2(x)= (f(x))^2$$.)

Thanks!

Oct 30, 2020

#1
+9142
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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

Oct 31, 2020
#2
+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