Question

Find the derivatives from the left and from the right at x = 1 (if they exist) of the following functions. Are the functions differentiable at x = 1?

`f(x) = sqrt(1 - x^2)`

Answer 1

`f(x) = sqrt(1 - x^2)` at x = 1

To find the left limit of `f(x)` at x = 1

Put x = 1 – h

h > 0

When x → 1

We have h → 0

`f"'"(1 - "h") =  lim_("h" -> 0) (f(1 - "h") - f(1))/(1 - "h" - 1)`

= `lim_("h" -> 0) (sqrt(1 - (1 - "h")^2) - sqrt(1 - 1^2))/(- "h")`

= `lim_("h" -> 0) (sqrt(1 - (1 - 2"h" + "h"^2)) - 0)/(- "h")`

= `lim_("h" -> 0) sqrt(1 + 1 + 2"h" - "h"^2)/(- "h")`

= `lim_("h" -> 0) sqrt("h"^2 (2/"h" - 1))/(- "h")`

= `lim_("h" -> 0) ("h" sqrt(2/"h" - 1))/(- "h")`

= `- lim_("h" -> 0) sqrt(2/"h" - 1)`

`f"'" (1^-) = - sqrt(2/0 - 1) = - sqrt(oo - 1)`

`f"'"(1^-) = - oo`  .........(1)

To find the right limit of `f(x)` at x = 1

Put x = 1 + h

h > 0

When x → 1

We have h → 0

`f"'"(1 + "h") =  lim_("h" -> 0) (f(1 + "h") - f(1))/(1 + "h" - 1)`

= `lim_("h" -> 0) (sqrt(1 - (1 + "h")^2) - sqrt(1 - 1^2))/"h"`

= `lim_("h" -> 0) sqrt(1 - (1 + 2"h" + "h"^2))/"h"`

= `lim_("h" -> 0) sqrt(1 - 1 - 2"h" - "h"^2)/"h"`

= `lim_("h" -> 0) sqrt(- "h"^2(1 + 2/"h"))/"h"`

= `lim_("h" -> 0) (sqrt(- 1) * "h")/"h" * sqrt(1 + 2/"h")`

= `lim_(h' -> 0) "i" * sqrt(1 + 2/"h")`

`f"'"(1^+) = "i" sqrt(1 + 2/0) = "i" xx oo`  .......(2)

From equations (1) and (2) we get

`f"'"(1^-)  ≠  f"'"(1^+)`

∴ `f"'"(x)` does not exist at x = 1

Hence `f(x) = sqrt(1 - x^2)` is n differentiable at x = 1.