AP Calculus - Graph

Question

State the intervals of the function f(x) = sqrt(x²-9) + sqrt(16-x²) is decreasing.

Answer 1

Domain of f(x)

f(x) = sqrt(x² – 9) + sqrt(16 – x²)

f(x) is undefined when |x| < 3 or |x| > 4

So the domain of x is limited to 3 ≤ |x| ≤ 4

which means –4 ≤ x ≤ –3 and 3 ≤ x ≤ 4

Where the derivative f’(x) = 0

f'(x) = x/sqrt(x² – 9) – x/sqrt(16 – x²)

f'(x) = 0 when sqrt(x² – 9) = sqrt(16 – x²)

which means x² – 9 = 16 – x²

which means 2x² = 16 + 9 = 25

which means x = ±5/√2 or |x| = 5/√2

The sign of the derivative f’(x)

f’(x) changes sign at 

x = –5/√2 on the interval –4 ≤ x ≤ –3

x = 5/√2 on the interval  3 ≤ x ≤ 4

When |x| > 5/√2 

sqrt(x² – 9)↑ and sqrt(16 – x²)↓ (where ↑ means increases and ↓ means decreases)

When |x| < 5/√2 

sqrt(x² – 9)↓ and sqrt(16 – x²)↑

So for f'(x) = x/sqrt(x² – 9) – x/sqrt(16 – x²)

the second term is dominant for |x| > 5/√2, and the first term is dominant for |x| < 5/√2

On –4 ≤ x ≤ –3 where both terms are negative, 

when |x| > 5/√2 this means f'(x) > 0 and f(x) is increasing 

when |x| < 5/√2 this means f'(x) < 0 and f(x) is decreasing

On 3 ≤ x ≤ 4 where both terms are positive,

when |x| > 5/√2 this means f'(x) < 0 and f(x) is decreasing

when |x| < 5/√2 this means f'(x) > 0 and f(x) is increasing

    

Answer

The intervals where f(x) is decreasing are

–5/√2 ≤ x ≤ –3 and 5/√2 ≤ x ≤ 4

In interval notation [–5/√2, –3] and [5/√2, 4]

Answer 2

Answer 3

Let f(x)=A(x)+B(x) where

1. A(x)=Sqrt(x^2-9); B(x)=Sqrt(16-x^2) (1)

From (1) it it follows that

1. A(x) & B(x) together (and thus f(x)) , can have real values only for x from the interval [3,4] (2a)
2. A(x)>=0; B(x)>=0 => f(x) >0 (2b)
3. f(x) in NOT a constant but at the ends of the [3,4] interval f(x) has the same value : f(3)=f(4)=Sqr(7) (2c)
4. from (2b) and (2c) it follows that f(x) must have at least one extremum inside the interval [3,4] (3)

Because A(x)^2+B(x)^2=cons=25 (4)

by taking the derivative of both sides of (4) we'd get

1. A(dA/dx)+B(dB/dx)=0 (5)

Using the condition of extremum of function f(x) , [ df/dx=0 ]

1. df/dx=dA/dx+dB/dx=0 (6) Combining it with (5), we 'd get
2. dA/dx ^, (A-B)=0 (7)

B/c dA/dx =x/Sqrt(x^2-9) cannot assume zero value for x from [3,4] , the only solution of (7) would be

1. A=B or Sqrt (x^2-9)=Sqrt(16-x^2) => (x^2-9)=(16-x^2) => 2x^2=25 (8)

which has a single solution x_m=5/Sqrt(2) (9)

(the other solution x_m= - 5/Sqrt(2) is negative thus - outside of the [3,4] interval)

[A side Note: It is interesting that at that point of extremum, A(x_m)=B(x_m)=Sqrt(7/2) (10) Indeed,

1. A(5/Sqrt(2))=Sqrt(25/2-9)=Sqrt(7/2) ; B(5/Sqrt(2)=Sqrt(16-25/2)=Sqrt(7/2) So, f(x_m)=2Sqrt(7/2)=Sqrt(14) ]

Because for x from [3,4] the function f(x)>0 ; f(3)=f(4); and f(x) has a single extremum in [3,4] at x=x_m that extremum (9) could only be a MAXIMUM. Therefore, the function f(x) would behave the following way:

1. Start at f(3)=Sqrt(7)
2. Increase between x=3 and x=x_m=5/Sqrt(2) reaching at x=x_m the maximal value of Sqrt(14)
3. From x=x_m to x=4 decrease reaching its starting value Sqrt(7) at x=4

[Side note 2: the point of max x_m=5/Sqrt(2) = approx 3.54 is just 0.04 above the middle point (3.5 of the interval [3.4] so, f(x) is slightly asymmetric relative to the middle of the interval [3,4] with its graph slightly bent to the right]

Hope that would be of help

Dr. Ariel B.