Answer:
y" = -24 / y³
Step-by-step explanation:
6x² + y² = 4
Take the derivative of both sides with respect to x.
12x + 2y y' = 0
Again, take the derivative of both sides with respect to x.
12 + 2y y" + y' (2y') = 0
12 + 2y y" + 2(y')² = 0
Solve for y' in the first equation.
2y y' = -12x
y' = -6x/y
Substitute and solve for y":
12 + 2y y" + 2(-6x/y)² = 0
12 + 2y y" + 2(36x²/y²) = 0
12 + 2y y" + 72x²/y² = 0
6y² + y³ y" + 36x² = 0
y³ y" = -36x² − 6y²
y" = (-36x² − 6y²) / y³
Solve for y² in the original equation and substitute:
y² = 4 − 6x²
y" = (-36x² − 6(4 − 6x²)) / y³
y" = (-36x² − 24 + 36x²) / y³
y" = -24 / y³
The given function f(x) = |x + 3| has both an absolute maximum and an absolute minimum.
What do you mean by absolute maximum and minimum ?
A function has largest possible value at an absolute maximum point, whereas its lowest possible value can be found at an absolute minimum point.
It is given that function is f(x) = |x + 3|.
We know that to check if function is absolute minimum or absolute maximum by putting the value of modulus either equal to zero or equal to or less than zero and simplify.
So , if we put |x + 3| = 0 , then :
± x + 3 = 0
±x = -3
So , we can have two values of x which are either -3 or 3.
The value 3 will be absolute maximum and -3 will be absolute minimum.
Therefore , the given function f(x) = |x + 3| has both an absolute maximum and an absolute minimum.
Learn more about absolute maximum and minimum here :
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Answer:
If your weekly allowance is 5, then for Christmas week your allowance was 15, then your allowance went up by: 5 =3 - 1 = 2 x 100 =200%.
Answer:
Use inverse cos to solve for x.
hope this helps
