# Complete the left-hand derivativef'_(a)= lim (f(a+h)-f(a))/h as h->0 from the left side and right-hand derivative f'+(a)= lim (f(a+h)-f(a))/h as h->0 from the right side of f at x=2 when...

Complete the left-hand derivative

f'_(a)= lim (f(a+h)-f(a))/h as h->0 from the left side

and right-hand derivative

f'+(a)= lim (f(a+h)-f(a))/h as h->0 from the right side

of f at x=2 when

f(x)= 3-x, x<2,

1/(3-x), x> or equal to 2.

Use your results to determine if f'(2) exists, and if it does, find its value.

### 1 Answer | Add Yours

You should evaluate the side limits to find if f'(2) exists such that:

`lim_(x->2,x<2)(3-x-f(2))/(x-2) = lim_(x->2,x<2)(3-x-(1/(3-2)))/(x-2)`

`lim_(x->2,x<2)(3-x-f(2))/(x-2) = lim_(x->2,x<2)(2-x)/(x-2)`

`lim_(x->2,x<2)(3-x-f(2))/(x-2) = lim_(x->2,x<2)-(x-2)/(x-2) = -1`

You need to evaluate right hand derivative such that:

`lim_(x->2,x>=2)(1/(3-x) - 1)/(x-2) = lim_(x->2,x>=2)(1 - 3 + x)/((3-x)(x-2))`

`lim_(x->2,x>=2)(1/(3-x) - 1)/(x-2) = lim_(x->2,x<2)(x-2)/((3-x)(x-2))`

`lim_(x->2,x>=2)(1/(3-x) - 1)/(x-2) = lim_(x->2,x<2)(1/(3-x))`

`lim_(x->2,x>=2)(1/(3-x) - 1)/(x-2) = 1/(3-2) = 1/1 = 1`

**Hence, since the left hand derivative and the right hand derivative have no equal values `f'_l(2) = -1!=1=f'_r(2), ` thus, the derivative of the function does not exist at `x=2` .**

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