4. compute the third derivative by taking the derivative of the second derivative.
5. determine which, if any, values of x found in step 3 do not make the third derivative zero.
6. find the corresponding values of y for the points that meet the above criteria and report the (x, y) points as points of inflection.
However, we must begin by stipulating that the denominator of the may not equal zero and restrict the values of x corresponding to that condition:
2x² - 7x - 4 ≠ 0
x ≠ 7/4 ± sqrt(81)/4
x ≠ 4, x ≠ -1/2
1. Compute the first derivative, using the quotient rule:
If f(x) = g(x)/h(x), then f'(x) = {g'(x)h(x) - g(x)h'(x)}/{h(x)}²
g(x) = 6x² + 5x - 4, then g'(x) = 12x + 5
h(x) = 2x² - 7x - 4, then h'(x) = 4x - 7
f'(x) = {(12x + 5)(2x² - 7x - 4) - (6x² + 5x - 4)(4x - 7)}/{2x² - 7x - 4}²
f'(x) = {(12x)(2x² - 7x - 4) + (5)(2x² - 7x - 4) + (6x² + 5x - 4)(-4x) + 7(6x² + 5x - 4)}/{2x² - 7x - 4}²
f'(x) = {24x³ - 84x² - 48x + 10x² - 35x - 20 - 24x³ - 20x² + 16x + 42x² + 35x - 20}/{2x² - 7x - 4}²
f'(x) = {-52x² - 32x - 40}/{2x² - 7x - 4}²
f'(x) = -4{13x² + 8x + 10}/{2x² - 7x - 4}²
2. Compute f"(x) using the same method:
{g'(x)h(x) - g(x)h'(x)}/{h(x)}²
g(x) = -4{13x² + 8x + 10}, g'(x) = -4{26x + 8}
h(x) = {2x² - 7x - 4}², h'(x) = 2{4x - 7}{2x² - 7x - 4}
This is going to get ugly:
{(-4{26x + 8})({2x² - 7x - 4}²) - (-4{13x² + 8x + 10})(2{4x - 7}{2x² - 7x - 4})}/{(2x² - 7x - 4)²}²
There is a common factor of {2x² - 7x - 4}/{2x² - 7x - 4} that can become 1:
{(-4{26x + 8})({2x² - 7x - 4}) - (-4{13x² + 8x + 10})(2{4x - 7})}/{2x² - 7x - 4}³
There is a common factor (-8) in the numerator:
f"(x) = -8{(13x + 4)(2x² - 7x - 4) - (13x² + 8x + 10)(4x - 7)}/{2x² - 7x - 4}³
3. Set the second derivative equal to zero and solve for values of x:
-8{(13x + 4)(2x² - 7x - 4) - (13x² + 8x + 10)(4x - 7)}/{2x² - 7x - 4}³ = 0
{(13x + 4)(2x² - 7x - 4) - (13x² + 8x + 10)(4x - 7)} = 0
{(13x + 4)(2x² - 7x - 4) + (-13x² - 8x - 10)(4x - 7)} = 0
26x³ - 91x² - 52x + 8x² - 28x - 16 - 52x³ - 32x² -40 + 91² + 56x + 70 = 0
I have run out of time before I must leave for work. I will try to return to this later, if you need more help.