# Use f(x)= 1/(x)^(1/2)+1)) to answer the following: a) use the difference quotient f'(x)= lim z-->x f(x)-f(z)/x-z to find the f'(x) b) what is the rate of change of f with respect to x when x...

Use f(x)= 1/(x)^(1/2)+1)) to answer the following:

a) use the difference quotient f'(x)= lim z-->x f(x)-f(z)/x-z to find the f'(x)

b) what is the rate of change of f with respect to x when x =9

c)find in slope-intercept form the equation of the line tangent to f(x) at x=1.

*print*Print*list*Cite

a) You need to use limit definition of the derivative such that:

`lim_(h-gt0) (f(x+h)-f(x))/(h)`

`lim_(h-gt0) (1/(sqrt(x+h) + 1) - 1/(sqrtx + 1))/h`

`lim_(h-gt0) ((sqrtx + 1 - sqrt(x+h) - 1)/((sqrt(x+h) + 1)*(sqrtx + 1)))/h`

`lim_(h-gt0) (sqrtx - sqrt(x+h))/(h*(sqrt(x+h) + 1)*(sqrtx + 1))`

`lim_(h-gt0) ((sqrtx - sqrt(x+h))(sqrtx+ sqrt(x+h)))/(h*(sqrtx+ sqrt(x+h))(sqrt(x+h) + 1)*(sqrtx + 1))`

`lim_(h-gt0) (x - x - h)/(h*(sqrtx+ sqrt(x+h))(sqrt(x+h) + 1)*(sqrtx + 1))`

`lim_(h-gt0) ( - h)/(h*(sqrtx+ sqrt(x+h))(sqrt(x+h) + 1)*(sqrtx + 1))`

`lim_(h-gt0) ( - 1)/((sqrtx+ sqrt(x+h))(sqrt(x+h) + 1)*(sqrtx + 1))`

Substituting 0 for h yields:

`lim_(h-gt0) ( - 1)/((sqrtx+ sqrt(x+h))(sqrt(x+h) + 1)*(sqrtx + 1)) = ( - 1)/((sqrtx+ sqrt(x+0))(sqrt(x+0) + 1)*(sqrtx + 1))`

`lim_(h-gt0) ( - 1)/((sqrtx+ sqrt(x+h))(sqrt(x+h) + 1)*(sqrtx + 1)) = - 1/((2sqrtx)(sqrt x+ 1)^2)`

**Hence, evaluating the derivative using limit definition yields `f'(x) = - 1/((2sqrtx)(sqrt x + 1)^2).` **

b) You need to evaluate f'(x) at x = 9 to find the rate of change of the function such that:

`f'(9) = -1/((2sqrt 9)(sqrt9 + 1)^2))`

**`f'(9) = -1/6*16 =gt f'(9) = -1/96` **

c) You need to write the equation of the tangent line to the graph of function at x = 1 such that:

`y - f(1) = f'(1)(x - 1)`

You need to evaluate f(1) and f'(1) such that:

`f(1) = 1/2`

`f'(1) = -1/8`

`y -1/2 = (-1/8)(x - 1)`

You need to remember the slope intercept form of equation of tangent line such that:

`y = mx + b`

Hence, you need to isolate y to the left side such that:

`y = 1/2 - x/8 + 1/8`

`y = -x/8 + 5/8`

**Hence, evaluating the slope intercept form of equation of tangent line to the graph of function at x = 1 yields `y = -x/8 + 5/8.` **