Anonymous
Anonymous asked in Science & MathematicsMathematics · 1 month ago

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

Relevance
  • 1 month ago
    Favourite 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

    :)>

  • 1 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

Still have questions? Get answers by asking now.