Post Graduate Maths Calculus

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¿.