Question

Use the limit definition to compute the derivative of the function f(x)=32/31−x at x=−16, and find an equation of the tangent line

Answer 1

The derivative of the function at
`x = -16` equals
`(32)/(2209)`.

The equation of the tangent line that passes through the point
`\left(-16, (32)/(47) \right)` is
`y = (32)/(2209)\cdot x +(2016)/(2209)`.

Explanation:

From Differential Calculus we remember the following definition of derivative,
`f'(x)`:

`f'(x) = \lim_(h \to 0) (f(x+h)-f(x))/(h)` (Eq. 1)

Where
`f(x+h)` is the function evaluated at
`x = x+h`, dimensionless.

If we know that
`f(x) = (32)/(31-x)`, then
`f(x+h)` is:

`f(x+h) = (32)/(31-x-h)` (Eq. 2)

Now we proceed to expand (Eq. 1):

`f'(x) = \lim_(h \to 0) ((32)/(31-x-h)-(32)/(31-x) )/(h)` (Eq. 1b)

`f'(x) = 32\cdot \lim_(h \to 0) (31-x-31+x+h)/(h\cdot (31-x-h)\cdot (31-x))`

`f'(x) = 32\cdot \lim_(h \to 0) (1)/((31-x-h)\cdot (31-x))`

`f'(x) = (32)/((31-x)^(2))` (Eq. 3)

And the derivative is evaluated at
`x = -16`:

`f'(-16) = (32)/((31+16)^(2))`

`f'(-16) = (32)/(2209)`

The derivative of the function at
`x = -16` equals
`(32)/(2209)`.

This value represents the slope of the tangent line that passes through
`x = -16`, and value of
`y` is now found:

`f(-16) = (32)/(47)`

The tangent line is represented by the following model:

`y = m\cdot x + b` (Eq. 4)

Where:

`m` - Slope, dimensionless.

`b` - y-Intercept, dimensionless.

`x` - Independent variable, dimensionless.

`y` - Dependent variable, dimensionless.

If we know that
`m = (32)/(2209)`,
`x = -16` and
`y = (32)/(47)`, the y-Intercept is:

`b = y-m\cdot x` (Eq. 4b)

`b = (32)/(47)-(32)/(2209)\cdot (-16)`

`b = (2016)/(2209)`

The equation of the tangent line that passes through the point
`\left(-16, (32)/(47) \right)` is
`y = (32)/(2209)\cdot x +(2016)/(2209)`.