Calculus - 2px + 5p = 300

Question

1.The price p (in $) and demand x for a product are related by the equation 2px + 5p = 300. If the price is decreasing at a rate of $2 per month when the price is $10, find the rate of change of the demand.

2. Given f (x) = (x^2 − 4x) ^3/4 . Find the intervals where f is increasing/decreasing, and the x-values of any local extrema of f.

Answer: Increasing (if any) on __________

Decreasing (if any) on _____________

Local max (if any) at x = __________________________; Local min (if any) at x = __________________________

3.Given f(x) = ( x^2 + 1)/2x. Find the intervals where f is increasing/decreasing, and the x-values of any local extrema of f.

Answer: Increasing (if any) on _____________________

Decreasing (if any) on ___________________

Local maximum (if any) at x = __________________

Local minimum (if any) at x = __________________

Answer 1

1)

Calculate "x" when p = 10

2(10)x + 5(10) = 300

20x + 50 = 300

20x = 250

x = 12.5

Plug in x = 12.5, p' = -2, p = 10:

x' = (-2)[2(12.5)+5]/(-2(10))

x' = (-2)(30)/(-20) = 3 units/month

2) f (x) = (x² − 4x)^(3/4)

Realize that this function DNE for values between 0 and 4 because values of "x' in that range produce imaginary numbers (4th root of a negative)

f '(x) = (3/4)(x² − 4x)^(-1/4)•(2x - 4)

f '(x) = [3(2x - 4)]/[4(x² − 4x)^(1/4)]

f '(x) =(6x - 12)/[4(x(x − 4))^(1/4)]

Find the x-values where f '(x) DNE:

set the denominator equal to zero to get x = 0 and x = 4

Find the x-values where f '(x) = 0

set the numerator equal to zero to get x = 2 however, the function DNE there so we can ignore this.

Put the critical points (0and 4) on a number line and try values on each interval:

f '(-1) = -3 so the function is decreasing on (-∞, 0)

f '(5) = 3 so the function is increasing on (4, ∞)

Since f(0) and f(4) = 0, these are the absolute minimums for the function and there are no absolute maximums.

3) f(x) = (x² + 1)/(2x)

Use the quotient rule, (u/v)' = (u'v - uv')/v² to get:

f '(x) = [(2x)(2x) - (x² + 1)(2)]/(4x²)

f '(x) = [4x² - 2x² - 2]/(4x²)

f '(x) = [2x² - 2]/(4x²)

f '(x) = (x² - 1)/(2x²)

Find the values of "x" where f '(x) DNE

Set the denominator equal to zero to get x = 0

Find the values of "x" where f '(x) = 0

Set the numerator equal to zero to get x = 1 and x = -1

Put these three critical points (x = -1, 0, and 1) on a number line and test values in each interval:

f '(-2) = 0.37 so the function is increasing on (-∞, -1)

f '(-0.5) = -1.5 so the function is decreasing on ( -1, 0)

f '(0.5) = -1.5 so the function is decreasing on ( 0, 1)

f '(2) = 0.37 so the function is increasing on (1, ∞)

Because of these, we can deduce there is a local max at x = -1 (plugging x = -1 into the function, the we get f(-1) = -1) and there is a local min at x = 1 (plugging x = 1 into the function, the we get f(1) = 1 )

Answer 2

Time for implicit differentiation! Try that, it works.

-- Cheers, --Mr. d.