Decreasing Function Example

Question 1.

Determine where [f(x)=x^{2}-4x+1] is increasing and decreasing.

Step-by-step solution:

1. [\dfrac{d}{dx}f(x)=2x-4].
2. Critical point: solve [2x-4=0] ⇒ [x=2].
3. Test intervals: [(-∞,2)] and [(2,∞)].
   • At [x=0]: [2(0)-4=-4 < 0] → derivative negative → decreasing.
   • At [x=3]: [2(3)-4=2 > 0] → derivative positive → increasing.
4. Conclusion: [f(x)] is decreasing on [(-\infty,2)] and increasing on [(2,\infty)]. At [x=2] the function has a local minimum.

Question 2.

Find where [f(x)=x^{3}] is increasing or decreasing.

Step-by-step solution:

1. [\dfrac{d}{dx}f(x)=3x^{2}].
2. Critical point: [3x^{2}=0] ⇒ [x=0].
3. Sign of derivative: [3x^{2} \ge 0] for all x; it equals 0 only at x=0.
   • For x<0 and x>0, derivative > 0.
4. Conclusion: [f(x)] is increasing on [(-\infty,0)] and on [(0,\infty)]; in fact it is strictly increasing on (-∞,∞) (derivative nonnegative and function increases between any two points).

Question 3.

Determine monotonicity of [f(x)=\ln x] on its domain.

Step-by-step solution:

1. Domain: [x>0].
2. [\dfrac{d}{dx}f(x)][=\dfrac{1}{x}].
3. For [x>0], [\dfrac{1}{x} > 0].
4. Conclusion: [f(x)] is strictly increasing on [(0,∞)].

Question 4.

Where is [f(x)=\dfrac{1}{x}] increasing or decreasing? (Domain: [x\ne 0])

Step-by-step solution:

1. [\dfrac{d}{dx}f(x) ][= -\dfrac{1}{x^{2}}].
2. Denominator [x^{2}] > 0 for x≠0, so derivative is always negative.
3. Conclusion: [f(x)] is decreasing on both intervals [(-∞,0)] and [(0,∞)].

Question 5.

Decide intervals of increase / decrease for [f(x)=\sin x] on [[0,2\pi]].

Step-by-step solution:

1. [\dfrac{d}{dx}f(x)=\cos x].
2. Solve [\cos x=0] on [0,2\pi] ⇒ [x=\tfrac{\pi}{2},\ \tfrac{3\pi}{2}].
3. Test intervals: [(0,π/2)], [(π/2,3π/2)], [(3π/2,2π)]:
   • At [x=\tfrac{\pi}{4}]: [\cos(\pi/4)][=\tfrac{\sqrt{2}}{2}>0] → increasing.
   • At [x=\pi]: [\cos\pi=-1<0] → decreasing.
   • At [x=\tfrac{7\pi}{4}]: [\cos(7\pi/4)=\tfrac{\sqrt{2}}{2}>0] → increasing.
4. Conclusion: [\sin x] is increasing on [\big(0,\tfrac{\pi}{2}\big)] and [\big(\tfrac{3\pi}{2},2\pi\big)], and decreasing on [\big(\tfrac{\pi}{2},\tfrac{3\pi}{2}\big)].

Question 6.

Find where [f(x)=e^{x}-x] is increasing.

Step-by-step solution:

1. [\dfrac{d}{dx}f(x)=e^{x}-1].
2. Solve [e^{x}-1=0] ⇒ [e^{x}=1] ⇒ [x=0].
3. Test intervals: (-∞,0) and (0,∞):
   • At [x=-1]: [e^{-1}-1 ][\approx 0.3679 -1 ][= -0.6321 <0] → decreasing.
   • At [x=1]: [e^{1}-1 ][\approx 2.7183 -1 ][= 1.7183 >0] → increasing.
4. Conclusion: [f(x)] is decreasing on [(-∞,0)] and increasing on [(0,∞)]. At [x=0] there is a local minimum.

7.

Determine where [f(x)=\dfrac{x^{2}}{x^{2}+1}] is increasing / decreasing.

Step-by-step solution:

1. Differentiate (quotient rule):
   [\dfrac{d}{dx}f(x)][=\dfrac{2x(x^{2}+1) – x^{2}(2x)}{(x^{2}+1)^{2}} ][= \dfrac{2x}{(x^{2}+1)^{2}}].
2. Solve [\dfrac{2x}{(x^{2}+1)^{2}=0}] ⇒ numerator [2x = 0 ⇒ x=0].
3. Sign: denominator positive; sign determined by 2x:
   • x<0 ⇒ derivative <0 (decreasing)
   • x>0 ⇒ derivative >0 (increasing)
4. Conclusion: decreasing on [(-∞,0)], increasing on [(0,∞)]. Minimum at [x=0].

Question 8.

Find intervals of monotonicity for [f(x)=\sqrt{x}] on its domain.

Step-by-step solution:

1. Domain: [x\ge 0].
2. [\dfrac{d}{dx}f(x)][=\dfrac{1}{2\sqrt{x}}] for x>0.
3. For x>0, derivative > 0; at x=0 derivative is infinite but function is increasing from right.
4. Conclusion: [f(x)] is strictly increasing on [(0,∞)]. (Also non-decreasing on [[0,∞]].)

9.

Where is [f(x)=x – \dfrac{1}{x}] increasing or decreasing? (Domain x≠0)

Step-by-step solution:

1. [\dfrac{d}{dx}f(x)][=1 + \dfrac{1}{x^{2}}].
2. Since [\dfrac{1}{x^{2}} > 0], derivative = 1 + positive > 1 ⇒ always positive for x≠0.
3. Conclusion: [f(x)] is strictly increasing on both [(-∞,0)] and [(0,∞)].

10.

Consider the implicit relation [x^{2}+y^{2}=16] and take the upper semicircle [y>0]. For y as a function of x, determine where y is decreasing.

Step-by-step solution:

1. Differentiate implicitly: [2x + 2y,\dfrac{dy}{dx}=0] ⇒ [\dfrac{dy}{dx} = -\dfrac{x}{y}].
2. On the upper semicircle [y] [= +√(16 – x^{2}) > 0] for x∈[(-4,4)].
3. Sign of derivative is -x/y → since y>0 sign = – sign(x):
   • For x>0 ⇒ derivative negative ⇒ y decreasing.
   • For x<0 ⇒ derivative positive ⇒ y increasing.
4. Conclusion: On the upper semicircle, [y(x)] is decreasing for x∈[(0,4)] and increasing for x∈[(-4,0)].