University Calculus 2: Uniform Continuity Homework Assignment

This assignment solution addresses several problems related to uniform continuity in Calculus 2. The solution begins by investigating the uniform continuity of f(x) = x^5ln(x) on the interval [1, 2], employing the mean value theorem and derivative analysis. It proceeds to examine the uniform continuity of f(x) = πx^2 on (-∞, ∞) and f(x) = x^2cos(1/x) on (0, 1]. The solution also investigates the uniform continuity of f(x) = ln(x^4 + 1) on (-∞, ∞) and provides an example of a function that is not differentiable on [0, ∞) but is uniformly continuous. Finally, it determines the uniform continuity of f(x) = x^3sin(1/x) on (0, ∞) and g(x) = e^(-x^2) on (-∞, ∞), using relevant theorems and definitions to support the analysis.

Running head: POST GRADUATE MATHS CALCULUS 1
Post Graduate Maths Calculus
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POST GRADUATE MATHS CALCULUS 2
A) Investigate if f(x) = x5Ln x is continuous on [1,2]
F(x) =x5 ln x
F:R→A is uniformly continuous if ∀ ε > 0 ∃ σ >0 such that ∀ x,y∈ A
With |x-y|¿ σ ,we have |f(x)-f(y)| ¿ ε .
(show that f’(x) =1, and use the mean value theorem)
x5 ln x
Uv u’v + v’u.
Using the product rule; u’v + v’u =5x ln x + x5
x = 5 x4 ln x + x4
Since f’(x) =5x4 ln x +x4, it follows that |f’(x)| ≤ ε 1
Now to prove that |f’(x)|≤| is continuous let ϵ >0.
We can letσ =ϵ .
For all distinct values of x and y in [1, 2], the mean value theorem says that there is a value c:
| (5 x4 ln x ¿−¿/x-y|=|1/c|≤1
Thus, whenever |x-y|<σ , we have
| (5x4 ln x ¿−( 5 y 4 ln y)∨¿|x− y|<σ =ε .
This implies that f(x) = x5 ln x is uniformly continuous on [1,2].

POST GRADUATE MATHS CALCULUS 3
b) Investigate if f(x) =πx2
is uniformly continuous on (-∞ , ∞¿
F(x) =πx2
is continuous on (-∞ , ∞).
We need to find the derivative first.
∂
∂ x π x2 = ∂
∂ x eln ( πx2) = ∂
∂ x ex2
ln(π)
ex2
ln ¿ =ex ln ¿( ∂
∂ x xln ¿)
By applying the chain rule with functions ex and xln ¿.
Therefore = ∂
∂ x πx2 =πx2 ln (π )
Since f’(x) =πx2 ln (π ). It follows that |f’(x)| ≤ 1
Now to prove that |f’(x)≤ 1| is continuous let ε > 0
We can letσ =ε.
For all values in ( −∞ , ∞) the mean value theorem says that there is a value C:
| πx2 ln (π ) - π y2 ln ( y)/x-y| =|1/c|≤1
Thus whenever |x-y|<σ ,it follows that
|πx2 ln ¿) - π y2 ln ( y)|<|x-y|<σ =ε
Showing that f(x) = πx2 is continuous on (−∞ , 2¿.

POST GRADUATE MATHS CALCULUS 4
c) Investigate if f(x) =cos x2 ¿) is uniformly continuous on (0,1).
F(x) = x2 cos ¿). (0, 1).
Assuming x2 is continuous (as it was shown in the interval)
X=0 implies for some ε > 0 ∃aσ > 0 : |x|<σ then x2<ε.
Since| cos ( 1
x ))|≤ 1, it follows that |x|< σ
This implies that |f(x)=cos x2 ¿) <ε
Hence , f is continuous at x=0 , as desired.
d) Investigate if f(x) = ln(x4 + 1) is uniformly continuous on (-∞, ∞).
Solution
Recall F: ℝ → is uniformly continuous if ∀ ε >¿0 ∃ J >0 such that ∀ x , y ∈ R with⃓ x − y⃓ < ε
We have ⃓ f(x)-f(y) ⃓ <ε
PType equation here .roof
Let ε > 0 chhose σ = ε
2 , ∈ [ −∞ , ∞ ]∧x> y
Then ∀ x , y ∈ R with⃓ x− y⃓ <σ
ln x < x And Ln y<y
Now

POST GRADUATE MATHS CALCULUS 5⃓
Ln (x4+1) - Ln (y4+1) ⃓ < ⃓ x-y⃓ <2σ⃓
Ln (x4+1) - Ln (y4+1) ⃓ < ⃓ x-y⃓ <2( ε
2 ¿⃓
Ln (x4+1) - Ln (y4+1) ⃓ < ⃓ x-y⃓ <ε
e) Give an example where f is not differentiable on [0, ∞), but is uniformly continuous.
Solution
An example function is 3
√ x
Let f(x) =x1/3, on [0,
∞ ¿ the function is uniformly continuous as is any continuous function on a closed∧¿
bounded interval in R
The derivative of the function is bounded on domain [-1,1]⃓
1/3x2/3⃓ <1/3 for every ⃓ x⃓>1
As a result for every x,y outside of [-1,1]
Using mean value theorem⃓
f(x)-f(y)⃓ =⃓ f' ´ε⃓ x-y⃓≤ 1
3⃓ x− y⃓ where ε is a suitable point between x∧ y .
It follows that f is uniformly continuous on the complement of [0, ∞ ¿
f) Determine if the functions are uniformly continuous on their domain:

POST GRADUATE MATHS CALCULUS 6
i) f : (0,∞) → R, f(x) = x3 sin( 1
x2 )
solution
recall
f:R isuniformly continuous if ∀ ε > 0∃ σ >0 such that ∀ x , y ∈ R with⃓ x− y⃓ <ε
we have ⃓ f(x)-f(y)⃓ <ε
let ε > 0 choose σ=ε x , y ∈ [ 0 ,∞ ]∧x > y
then ∀ x , y ∈ R with⃓ x− y⃓ < σ
we have ⃓
x3sin [ 1
x2 ] − y 3 sin [ 1
y2 ]⃓ ≤⃓ x− y⃓ < σ=ε ≤⃓ x− y⃓ < ε
ii) g : (−∞,∞) → R+, g(x) = e-x^2
g:R isuniformly continuous if ∀ ε > 0∃ σ >0 such that ∀ x , y ∈ R with⃓ x− y⃓ <ε
we have ⃓ g(x)-g(y)⃓ <ε
let ε < 0 choose σ=ε x , y ∈ ( −∞ , ∞ ) ∧x> y
Then∀ x , y ∈ R with⃓ x− y⃓ < σ⃓
e− x2
−e− y2⃓ ≤⃓ x− y⃓ <σ =ε ≤⃓ x− y⃓ <ε