Respuesta :
Answer:
1
Step-by-step explanation:
Given function:
y = px² - 4x + q
To find the points of maxima or minima
[tex]\frac{dy}{dx}[/tex] = 0
or
2px - 4 + 0 = 0
or
px = 2
or
x = [2 ÷ p]
at p = 2
x = 2 ÷ 2 = 1
To check for minima
[tex]\frac{d^2y}{dx^2}[/tex] > 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
