Question

To find the equation of a line, we need the slope of the line and a point on the line. Since we are requested to find the equation of the tangent line at the point (36, 6), we know that (36, 6) is a point on the line. So we just need to find its slope. The slope of a tangent line to f(x) at x

Answer 1

`m = (1)/(12)`

Explanation:

Given

`(x,y) = (36,6)`

`f(x) = \sqrt x` ----- the equation of the curve

Required

The slope of f(x)

The slope (m) is calculated using:

`m = \lim_(h \to 0) (f(a + h) - f(a))/(h)`

`(x,y) = (36,6)` implies that:

`a = 36; f(a) = 6`

So, we have:

`m = \lim_(h \to 0) (f(a + h) - f(a))/(h)`

`m = \lim_(h \to 0) (f(36 + h) - 6)/(h)`

If
`f(x) = \sqrt x`; then:

`f(36 + h) = √(36 + h)`

So, we have:

`m = \lim_(h \to 0) (√(36 + h) - 6)/(h)`

Multiply by:
`√(36 + h) + 6`

`m = \lim_(h \to 0) ((√(36 + h) - 6)(√(36 + h) + 6))/(h(√(36 + h) + 6))`

Expand the numerator

`m = \lim_(h \to 0) (36 + h - 36)/(h(√(36 + h) + 6))`

Collect like terms

`m = \lim_(h \to 0) (36 - 36+ h )/(h(√(36 + h) + 6))`

`m = \lim_(h \to 0) (h )/(h(√(36 + h) + 6))`

Cancel out h

`m = \lim_(h \to 0) (1)/(√(36 + h) + 6)`

`h \to 0` implies that we substitute 0 for h;

So, we have:

`m = (1)/(√(36 + 0) + 6)`

`m = (1)/(√(36) + 6)`

`m = (1)/(6 + 6)`

`m = (1)/(12)`

Hence, the slope is 1/12