Compute the derivative dy/dx using the power, product, and chain rules. Given
x³ + y³ = 11xy
differentiate both sides with respect to x to get
3x² + 3y² dy/dx = 11y + 11x dy/dx
Solve for dy/dx :
(3y² - 11x) dy/dx = 11y - 3x²
dy/dx = (11y - 3x²)/(3y² - 11x)
The tangent line to the curve is horizontal when the slope dy/dx = 0; this happens when
11y - 3x² = 0
or
y = 3/11 x²
(provided that 3y² - 11x ≠ 0)
Substitute y into into the original equation:
x³ + (3/11 x²)³ = 11x (3/11 x²)
x³ + (3/11)³ x⁶ = 3x³
(3/11)³ x⁶ - 2x³ = 0
x³ ((3/11)³ x³ - 2) = 0
One (actually three) of the solutions is x = 0, which corresponds to the origin (0,0). This leaves us with
(3/11)³ x³ - 2 = 0
(3/11 x)³ - 2 = 0
(3/11 x)³ = 2
3/11 x = ³√2
x = (11•³√2)/3
Solving for y gives
y = 3/11 x²
y = 3/11 ((11•³√2)/3)²
y = (11•³√4)/3
So the only other point where the tangent line is horizontal is ((11•³√2)/3, (11•³√4)/3).
Hello :
f(x) = <span>a(x-k)²+k....... the point vertex is (h , k )
in this exercice : h = -3 and k = -1 s0 f(x) = a(x+3)²-1
find : a
but f(0)=8
a(0+3)² -1 = 8
9a-1=8
9a =9
a=1
f(x) =(x+3)²-1 =x²+6x+9-1
f(x) = x²+6x+8 .... (</span><span>the quadratic function in standard form )</span>
Speed = Distance ÷ time
Speed upstream = 84 ÷ 7 = 12 km/h
Speed downstream = 84 ÷ 3 = 28 km/h
Speed in still water = 1/2(28 + 12) = 20 km/h
Answer: 20 km/h
