Calculus Questions - Desklib

Calculus Questions
Table of Contents
Table of Contents.............................................................................................................................1
QUESTION 1..................................................................................................................................2
(a).................................................................................................................................................2
(b).................................................................................................................................................2
(c).................................................................................................................................................3
(d).................................................................................................................................................3
QUESTION 2..................................................................................................................................4
QUESTION 3..................................................................................................................................4
(a) Range......................................................................................................................................4
(b) Strictly decreasing interval.....................................................................................................5
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QUESTION 1
(a)
f(x) = (x9.5 – 3 + 1) (5 – [1 / (x2+1)])
Solution
f(x) = (x9.5 – 3 + 1) (5 – [1 / (x2+1)])
= 5 x6.5 - x6.5 / x3 + 5 – 1/x3
= 5 x6.5 – x3.5 + 5 – 1/x3
First derivative = f’(x)
f’(x) = 32.5 x5.5 – 3.5 x2.5 + 3x-4
Second derivative = f ’’(x) = 178.75 x4.5 – 8.75 x1.5 - 12 x-5
(b)
f(x) = exp √[x2 – (2/x) ]
Solution
f(x) = exp √[x2 – (2/x) ]
Let √[x2 – (2/x) ] = w
d/dx (ew) = dw/dx*ew
f’(x) = d/dx (√[x2 – (2/x) ]) * exp √[x2 – (2/x) ]
Again d/dx (√x) = 0.5 (x)-0.5
Applying this chain rule:
d/dx (√[x2 – (2/x) ]) = (2x + 2x-2)0.5 * [x2 – (2/x) ]-0.5
Thus f’(x) =(2x + 2x-2)0.5 * [x2 – (2/x) ]-0.5 * exp √[x2 – (2/x) ]
First derivative = 0.5*[x2 – (2/x) ]-0.5)* exp √[x2 – (2/x) ] * [2x + 2x-2]
Second derivative =
f’’(x) = d/dx [(0.5) (x2 –2x-1)-0.5 exp √[x2 – (2/x) ] * [2x + 2x-2]]
Let u = [(0.5) (x2 –2x-1)-0.5 * [2x + 2x-2] ]
On simplifying we have u = 2x3 + 1 -2 – 2x-3
du/dx = 3x2 + 6x-4
Let v = exp √[x2 – (2/x) ]
dv/dx = 1 / {2 *√ [x2 – (2/x) ] } * (2x + 2x-2) exp √[x2 – (2/x) ]
so f ’’(x) = u dv/dx + v* du/dx
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