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).
Answer:
f(x)^-1 = x/4
Step-by-step explanation:
f(x)=4x
inverse
x = 4y
rewrite
4y = x
y = x/4
Answer:
f(x)^-1 = x/4
You would input 10 as m into the equation
t= 0.05(10) +60
t=0.5+60
t=60.5
Answer:
The lines representing these equations intercept at the point (-4,2) on the plane.
Step-by-step explanation:
When we want to find were both lines intercept, we are trying to find a pair of values (x,y) that belongs to both equations, which means that it satisfies both equations at the same time.
Therefore, we can use the second equation that gives us the value of y in terms of x, to substitute for y in the first equation. Then we end up with an equation with a unique unknown, for which we can solve:
Next we use this value we obtained for x (-4) in the same equation we use for substitution in order to find which y value corresponds to this:
Then we have the pair (x,y) that satisfies both equations (-4,2), which is therefore the point on the plane where both lines intercept.
x^2-9x-6=0
Use the quadratic formula to get (9+sqrt(105))/2 or (9-sqrt(105))/2
