The local extremas are

Question

y=(4x^2-9x)/2x+6

Answer 1

To find local extrema (local minima or maxima), you'll first need to take the derivative of the function. We'll use the Quotient Rule:

 

( (f(x)/g(x) )' = (g(x)f'(x) - f(x)g'(x)) / (g²(x)), where f(x) = function's numerator and g(x) = function's denominator

 

For your function:

 

f(x) = 4x² - 9x

f'(x) = 2(4x) - 9 = 8x - 9

g(x) = 2x+6

g'(x) = 2

 

So, applying the rule and simplifying gives you:

 

y' = ((2x+6)(8x-9) - (4x²-9x)(2)) / (2x+6)²

y' = ((16x² - 18x + 48x - 54) - (8x² - 18x)) / ((2x+6)(2x+6))

y' = (16x² + 30x - 54 - 8x² + 18x) / ((2x+6)(2x+6))

y' = (8x² + 48x - 54) / ((2x+6)(2x+6))

 

Factor a 2 out of the numerator and a 2 out of the denominator, which will cancel out:

 

y' = (2(4x² + 24x - 27)) / (2(x+3)(2x+6))

y' = (4x² + 24x - 27) / ((x+3)(2x+6))

 

 

Now, we'll let y' = 0 and solve for the zeroes of the derivative using the Quadratic Formula. Each zero is a single critical point of the original function.

 

0 = (4x² + 24x - 27) / ((x+3)(2x+6))

0 = 4x² + 24x - 27

 

x = (-24 ± √(24² - 4(4)(-27))) / 2(4)

x = (-24 ± √(576 + 432)) / 8

x = (-24 ± √1008) / 8

x = (-24 ± 31.75) / 8

x = (-24 + 31.75)/8   OR   (-24 - 31.75)/8

x = 7.75/8   OR   -55.75/8

x = 0.97   OR   -6.97

 

So, there are critical points at x = 0.97 and x = -6.97. However, not all critical points are extrema. To test whether each of these points is an extremum, we'll need to find out what y' is like to the left and right of each point. So, we'll plug test values for x into y' to find out where y' is negative and where it's positive.

 

We want to know what y' is like to the left of -6.97, between -6.97 and 0.97, and to the right of 0.97. So, plug in x = -8, x = 0, and x = 2.

 

x = -8:

y' = (4(-8)² + 24(-8) - 27) / (((-8)+3)(2(-8)+6))

y' = (4(64) + (-192) - 27) / ((-5)(-16+6))

y' = (256 - 219) / ((-5)(-10))

y' = 37/50

 

Left of x = -6.97, y' is positive.

 

x = 0:

y' = (4(0)² + 24(0) - 27) / (((0)+3)(2(0)+6))

y' = (0 + 0 - 27) / ((3)(0 + 6))

y' = -27 / ((3)(6))

y = -27/18 = -3/2

 

Between x = -6.97 and x = 0.97, y' is negative.

 

x = 2:

y' = (4(2)² + 24(2) - 27) / (((2)+3)(2(2)+6))

y' = (4(4) + 48 - 27) / ((5)(4 + 6))

y' = (16 + 21) / ((5)(10))

y' = 37/50

 

Right of x = 0.97, y' is positive.

 

 

So, y' looks like this:

 

 

...----------positive------------|-6.97|---------negative--------|0.97|--------positive----------...

 

 

Since y' switches signs at both x = -6.97 and x = 0.97, both values are extrema of the original function. Now all that remains is to find their corresponding y-coordinates by plugging each x into the original function and solving for y:

 

x = -6.97:

y = (4(-6.97)² - 9(-6.97)) / (2(-6.97)+6)

y = (4(48.58) + 62.73) / (-13.94+6)

y = (194.32 + 62.73) / (-7.94)

y = 257.05/(-7.94)

y = -32.37

 

x = 0.97:

y = (4(0.97)² - 9(0.97)) / (2(0.97)+6)

y = (4(0.94) - 8.73) / (1.94+6)

y = (3.76 - 8.73) / (7.94)

y = -4.97/7.94

y = -0.63

 

The original function's local extrema are: (-6.97, -32.37) and (0.97, -0.63)