Question

For the function y = (x² + 9) / (x + 15) at (3, 1), find the following. (Give exact answers. Do not round.) (a) the slope of the tangent line (b) the instantaneous rate of change of the function

Answer 1

The slope of the tangent line at the point (3, 1) is 1, and the instantaneous rate of change of the function at that point is also 1.

Step-by-step explanation:

To find the slope of the tangent line at the point (3, 1), we need to find the derivative of the function. First, we can simplify the function by dividing the numerator by the denominator, which gives us y = x - 6. Taking the derivative of y with respect to x, we get dy/dx = 1. So the slope of the tangent line is 1.

The instantaneous rate of change of a function is given by the derivative. In this case, since the derivative of y is 1, the instantaneous rate of change of the function at the point (3,1) is also 1.

Answer 2

The slope of the tangent line to the function
`$y=(x^2+9)/(x+15)$` at the point (3,1) is
`$(5)/(18)$` , this means that for a small change in x near x=3, the function y changes by
`$(5)/(18)$`.

To find the slope of the tangent line to the function
`$y=(x^2+9)/(x+15)$` at the point
`$(3,1)$`, we can use the concept of differentiation.

(a) To find the slope of the tangent line, we need to find the derivative of the function with respect to
`$x$`. The derivative of the function
`$y$` is denoted as
`$(dy)/(dx)$` or
`$y'$`.

To find the derivative, we can use the quotient rule:
`$(d)/(dx)\left((u)/(v)\right) = (v(du)/(dx) - u(dv)/(dx))/(v^2)$`, where u and v are functions of x.

Applying the quotient rule to our function, we have:

`$y'=((x+15)(2x) - (x^2+9)(1))/((x+15)^2)$`

Simplifying this expression, we get:

`$y'=(2x^2+30x-1x^2-9)/((x+15)^2)$`

`$y'=(x^2+30x-9)/((x+15)^2)$`

To find the slope of the tangent line at (3,1), we substitute x=3 into y':

`$y'=(3^2+30(3)-9)/((3+15)^2)$`

`$y'=(9+90-9)/(18^2)$`

`$y'=(90)/(324)$`

Simplifying the fraction, we have:

`$y'=(5)/(18)$`

Therefore, the slope of the tangent line to the function
`$y=(x^2+9)/(x+15)$` at the point (3,1) is
`$(5)/(18)$`.

(b) The instantaneous rate of change of the function is equal to the slope of the tangent line. Therefore, the instantaneous rate of change of the function at the point (3,1) is also
`$(5)/(18)$`. This means that for a small change in x near x=3, the function y changes by
`$(5)/(18)$`.