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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

:)>

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

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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