# Rate of Change!!?

NEED HELP WITH THESE!!!!!

Determine the slope of the line that is tangent to the graph of each function at the given point. Then determine the value of x at which there is not tangent line.

a) f (x) = -5x / (2x + 3) , where x = 2

b) f (x) = (2x^2 - 6x) / (3x + 5) , where x = -2

### 2 Answers

- Wayne DeguManLv 71 month agoFavourite answer
With rational functions, the denominator cannot be zero.

i.e. 2x + 3 ≠ 0

so, x = -3/2...not a tangent line

Also, 3x + 5 ≠ 0

so, x = -5/3...not a tangent line

Using the quotient rule we have:

a) f '(x) = [-5(2x + 3) - 2(-5x)]/(2x + 3)²

i.e. -15/(2x + 3)²

At x = 2 we have:

f '(2) = -15/(7)² => -15/49

b) f '(x) = [(4x - 6)(3x + 5) - 3(2x² - 6x)]/(3x + 5)²

i.e. 2(3x² + 10x - 15)/(3x + 5)²

At x = -2 we have:

f '(-2) = 2(-23)/(-1)² => -46

:)>

- micatkieLv 71 month ago
a)

f'(x)

= [(2x + 3)(-5x)' - (-5x)(2x + 3)'] / (2x + 3)²

= [(2x + 3)(-5) - (-5x)(2)] / (2x + 3)²

= [-10x - 15 + 10x] / (2x + 3)²

= -15 / (2x + 3)²

Slope of the tangent at (x = 2)

= f'(x) at (x = 2)

= -15 / (2*2 + 3)²

= -15/49

f(x) = -5x/(2x + 3)

When x = -3/2, denominator 2x + 3 = 0, and thus f(-3/2) is undefined.

Hence, there is no tangent at x = -3/2

====

b)

f'(x)

= [(3x + 5)(2x² - 6x)' - (2x² - 6x)(3x + 5)'] / (3x + 5)²

= [(3x + 5)(4x - 6) - (2x² - 6x)(3)] / (3x + 5)²

= [12x² - 18x + 20x - 30 - 6x² + 18] / (3x + 5)²

= (6x² + 2x - 12) / (3x + 5)²

Slope of the tangent at (x = -2)

= f'(x) at (x = -2)

= [6(-2)² + 2(-2) - 12] / [3(-2) + 5)²

= 8

f(x) = (2x² - 6x)/(3x + 5)

When x = -5/3, denominator 3x + 5 = 0, and thus f(-5/3) is undefined.

Hence, there is no tangent at x = -5/3