Find the value of k such that f(x) is differentiable at x=1. Please help!

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

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 .