Question

The function y=px2−4x+qy=px2−4x+q in the x-y plane attains a minimum value. What is the value of xx? 1) p=2p=2 2) q=5

Answer 1

Answer:

1

Step-by-step explanation:

Given function:

y = px² - 4x + q

To find the points of maxima or minima

$\frac{dy}{dx}$ = 0

or

2px - 4 + 0 = 0

or

px = 2

or

x = [2 ÷ p]

at p = 2

x = 2 ÷ 2 = 1

To check for minima

$\frac{d^2y}{dx^2}$ > 0

thus,

2p - 0 > 0

2p > 0

for positive value of p function is minimum at x = 1

2) at q = 5 also x = 1 as value of x is independent of q