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Math  /  Calculus

QuestionFind any relative extrema of the function. (Round your ar f(x)=arctan⁡(x)−arctan⁡(x−9)f(x)=\arctan (x)-\arctan (x-9) relative maximum (x,y)=(4.5,2.702x)(x, y)=(4.5,2.702 x)

Studdy Solution

STEP 1

1. The function given is f(x)=arctan⁡(x)−arctan⁡(x−9) f(x) = \arctan(x) - \arctan(x-9) .
2. Relative extrema occur where the derivative is zero or undefined.
3. We need to find the derivative of the function and solve for critical points.
4. Evaluate the second derivative to determine the nature of the critical points.

STEP 2

1. Find the first derivative of the function.
2. Solve for critical points where the first derivative is zero or undefined.
3. Use the second derivative test to determine the nature of the critical points.
4. Calculate the function value at the critical points to find relative extrema.

STEP 3

Find the first derivative of the function f(x)=arctan⁡(x)−arctan⁡(x−9) f(x) = \arctan(x) - \arctan(x-9) .
The derivative of arctan⁡(x) \arctan(x) is 11+x2 \frac{1}{1+x^2} .
So, the derivative of the function is:
f′(x)=11+x2−11+(x−9)2 f'(x) = \frac{1}{1+x^2} - \frac{1}{1+(x-9)^2}

STEP 4

Solve for critical points where the first derivative is zero or undefined.
Set the derivative equal to zero:
11+x2−11+(x−9)2=0 \frac{1}{1+x^2} - \frac{1}{1+(x-9)^2} = 0
Simplify and solve for x x :
11+x2=11+(x−9)2 \frac{1}{1+x^2} = \frac{1}{1+(x-9)^2}
Cross-multiply to get:
1+(x−9)2=1+x2 1 + (x-9)^2 = 1 + x^2
Simplify:
(x−9)2=x2 (x-9)^2 = x^2
x2−18x+81=x2 x^2 - 18x + 81 = x^2
−18x+81=0 -18x + 81 = 0
18x=81 18x = 81
x=8118=4.5 x = \frac{81}{18} = 4.5

STEP 5

Use the second derivative test to determine the nature of the critical points.
Find the second derivative of f(x) f(x) :
f′′(x)=−2x(1+x2)2+2(x−9)(1+(x−9)2)2 f''(x) = -\frac{2x}{(1+x^2)^2} + \frac{2(x-9)}{(1+(x-9)^2)^2}
Evaluate the second derivative at x=4.5 x = 4.5 :
f′′(4.5)=−2(4.5)(1+(4.5)2)2+2(4.5−9)(1+(4.5−9)2)2 f''(4.5) = -\frac{2(4.5)}{(1+(4.5)^2)^2} + \frac{2(4.5-9)}{(1+(4.5-9)^2)^2}
f′′(4.5)=−9(1+20.25)2+−9(1+20.25)2 f''(4.5) = -\frac{9}{(1+20.25)^2} + \frac{-9}{(1+20.25)^2}
f′′(4.5)=−18(1+20.25)2 f''(4.5) = -\frac{18}{(1+20.25)^2}
Since f′′(4.5)<0 f''(4.5) < 0 , the function has a relative maximum at x=4.5 x = 4.5 .

STEP 6

Calculate the function value at the critical point to find the relative maximum.
f(4.5)=arctan⁡(4.5)−arctan⁡(4.5−9) f(4.5) = \arctan(4.5) - \arctan(4.5-9)
f(4.5)=arctan⁡(4.5)−arctan⁡(−4.5) f(4.5) = \arctan(4.5) - \arctan(-4.5)
Using the property arctan⁡(−x)=−arctan⁡(x) \arctan(-x) = -\arctan(x) :
f(4.5)=arctan⁡(4.5)+arctan⁡(4.5) f(4.5) = \arctan(4.5) + \arctan(4.5)
f(4.5)=2arctan⁡(4.5) f(4.5) = 2\arctan(4.5)
Using a calculator, approximate arctan⁡(4.5)≈1.351 \arctan(4.5) \approx 1.351 :
f(4.5)≈2×1.351=2.702 f(4.5) \approx 2 \times 1.351 = 2.702
The relative maximum is at (x,y)=(4.5,2.702) (x, y) = (4.5, 2.702) .
The relative maximum of the function is at (x,y)=(4.5,2.702) (x, y) = (4.5, 2.702) .

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