Example Set 1 – Increasing and Decreasing Function

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

Step-by-step Solution:

Compute derivative: [\dfrac{d}{dx}f(x)=2x-4].
Find critical point: solve [2x-4=0] ⇒ [x=2].
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.
Conclusion: [f(x)] is decreasing on [(-\infty,2)] and increasing on [(2,\infty)]. (Local minimum at [x=2].)

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

Step-by-step Solution:

Derivative: [\dfrac{d}{dx}f(x)=3x^{2}].
Solve [3x^{2}=0] ⇒ [x=0].
Sign: [3x^{2} \ge 0] for all x; it is 0 at [x=0] and positive elsewhere.
Conclusion: [f(x)] is increasing on [(-\infty,0)] and increasing on [(0,\infty)]; in fact [x^{3}] is strictly increasing on [(-\infty,\infty)] (derivative nonnegative and function increases between any two points).

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

Step-by-step Solution:

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

Step-by-step Solution:

Derivative: [\dfrac{d}{dx}f(x) ][= -\dfrac{1}{x^{2}}].
Since [x^{2}>0] for x≠0, derivative is always negative.
Conclusion: [f(x)] is decreasing on [(-\infty,0)] and decreasing on [(0,\infty)].

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

Step-by-step Solution:

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

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

Step-by-step Solution:

Derivative: [\dfrac{d}{dx}f(x)=e^{x}-1].
Solve [e^{x}-1=0] ⇒ [x=0].
Test:
- For [x<0] (e.g., [x=-1]): [e^{-1}-1<0] ⇒ decreasing.
- For [x>0] (e.g., [x=1]): [e^{1}-1>0] ⇒ increasing.
Conclusion: [f(x)] is decreasing on [(-\infty,0)] and increasing on [(0,\infty)].

Determine monotonicity of [f(x)=\dfrac{x^{2}}{x^{2}+1}].

Step-by-step Solution:

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}}].
Solve [2x/(1+x^{2})^{2}=0] ⇒ [x=0].
Sign: numerator [2x] decides sign.
- [x<0] ⇒ derivative <0 ⇒ decreasing.
- [x>0] ⇒ derivative >0 ⇒ increasing.
Conclusion: [f(x)] is decreasing on [(-\infty,0)] and increasing on [(0,\infty)].

Find where [f(x)=\sqrt{x}] is increasing on its domain.

Step-by-step Solution:

Domain: [x\ge 0].
Derivative for x>0: [\dfrac{d}{dx}f(x)][=\dfrac{1}{2\sqrt{x}}] which is >0 for x>0.
At x=0 derivative is not finite, but function increases to the right.
Conclusion: [f(x)] is strictly increasing on [(0,\infty)] (and non-decreasing on [[0,\infty]]).

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

Step-by-step Solution:

Derivative: [\dfrac{d}{dx}f(x)][=1+\dfrac{1}{x^{2}}].
Since [\dfrac{1}{x^{2}}>0], [1+\dfrac{1}{x^{2}}>1>0] for all x≠0.
Conclusion: [f(x)] is strictly increasing on [(-\infty,0)] and strictly increasing on [(0,\infty)].

Implicit curve: [x^{2}+y^{2}=16]. For the upper semicircle [y>0], determine where y (as a function of x) is decreasing.

Step-by-step Solution:

Differentiate implicitly: [2x+2y,\dfrac{dy}{dx}=0] ⇒ [\dfrac{dy}{dx}=-\dfrac{x}{y}].
On the upper semicircle [y] [= +\sqrt{16-x^{2}} > 0] for x∈[(-4,4)].
Sign of derivative is [-x/y], with y>0:
- For [x>0] ⇒ [-x/y <0] ⇒ y decreasing.
- For [x<0] ⇒ [-x/y >0] ⇒ y increasing.
Conclusion: On the upper semicircle y(x) is decreasing for x∈[(0,4)] and increasing for x∈[(-4,0)].