what is the maximum distance between f and g when f(x)=1/2x^2 and g(x)=1/16x^4-1/2x^2 on closed interval [0,4]
let me know if you want the maximum vertical distance between them.
If you want the distance such that they have a common normal at that place, the solution appears to be a nightmare.
Reiny, i need the maximum vertical distance between the two functions of the graph.
ahh, not so hard
Did you make a rough sketch?
The two curves intersect at x=0 and x=4, which just happens to be the domain to consider.
vertical height = x^2/2 - (x^4/16 - x^2/2)
= x^2 - x^4/16
d(height)/dx = 2x - 4x^3/16 = 2x - x^3/4
= 0 for max/min of height
x(2 - x^2/4) = 0
x = 0 , which would give a min height
or
2- x^2/4 = 0
x^2 = 8
x = √8 or 2√2
max distance = x^2 - x^4/16
= 8 - 64/16 = 8-4 = 4 units
check my arithmetic
To find the maximum distance between two functions, we first need to calculate the distance between them at each point within the given interval. Then, we find the maximum value among all those distances.
Let's start by calculating the distance between the two functions at each point, x, within the interval [0,4]. The distance between two functions f(x) and g(x) at a specific point x is given by |f(x) - g(x)|.
Function f(x) = 1/2x^2, and function g(x) = 1/16x^4 - 1/2x^2. Therefore, the distance at any point x is |f(x) - g(x)| = |(1/2x^2) - (1/16x^4 - 1/2x^2)|.
Simplifying this expression, we have |(1/2x^2) - (1/16x^4 - 1/2x^2)| = |1/16x^4 - 1/2x^2|.
Now, we'll compute the distances at the endpoints and critical points within the interval [0,4].
1. At x = 0:
|1/16(0)^4 - 1/2(0)^2| = 0
2. At x = 4:
|1/16(4)^4 - 1/2(4)^2| = |1/2 - 8| = 7.5
3. To find any critical points, we differentiate the expression 1/16x^4 - 1/2x^2 and solve for x when the derivative is equal to zero.
Differentiating 1/16x^4 - 1/2x^2 with respect to x yields:
d/dx (1/16x^4 - 1/2x^2) = 1/4x^3 - x
Setting the derivative equal to zero:
1/4x^3 - x = 0
x(1/4x^2 - 1) = 0
This equation is satisfied when x = 0 and when 1/4x^2 - 1 = 0.
For 1/4x^2 - 1 = 0:
1/4x^2 = 1
x^2 = 4
x = ±2
Therefore, we have critical points at x = 0 and x = ±2.
Now, let's substitute each critical point and the endpoints into |1/16x^4 - 1/2x^2| to find the distance:
1. At x = 0:
|1/16(0)^4 - 1/2(0)^2| = 0
2. At x = ±2:
|1/16(±2)^4 - 1/2(±2)^2| = |1/4 - 2| = 1.75
3. At x = 4:
|1/16(4)^4 - 1/2(4)^2| = |1/2 - 8| = 7.5
The maximum distance between f and g within the interval [0,4] is 7.5.
Therefore, the answer is 7.5.