Calculus Questions - Desklib

This article contains solutions to calculus questions on topics like derivatives, range, and strictly decreasing interval. The solutions are explained step-by-step for better understanding. The article is relevant for students studying calculus in college or university.

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
1

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
2

On substituting values, we have:
Second derivative:
f ’’(x) = [(0.5) (x2 –2x-1)* [2x + 2x-2] ] 1 / {2 *√ [x2 – (2/x) ] } * (2x + 2x-2) exp √[x2 – (2/x) ]
+ exp √[x2 – (2/x) ] [3x2 + 6x-4 ]
(c)
f(x) = [1/ 5x +4] exp (3 – 3x2)
Solution
First derivative:
Let [1/ 5x +4] = u and exp (3 – 3x2) = v
d (uv) = uv’ + u’v where u’ and v’ are first derivative of function u and v
u = [1/ 5x +4]
u’ = du/dx = -5 / [5x +4]2
v = exp (3 – 3x2)
v’ = dv/dx = -6x * exp (3 – 3x2)
f’(x) = uv’ + u’v
On substituting values:
f’(x) = [1/ 5x +4] [-6x * exp (3 – 3x2)] + {-5 / [5x +4]2 * exp (3 – 3x2)}
f’(x) = = { exp (3 – 3x2) / [5x + 4] } * {(-6x- 5) / [5x + 4] }
(d)
f(x) = [6x2 + 2] (x+1)
Solution
Using the derivative formula:
If a and b are function of x and in the form (ab ) then they can be simplified as: (ab ) = eb ln a
From observation it can be seen that a = [6x2 + 2] and b = (x+1) so
f(x) = [6x2 + 2] (x+1) = e(x+1) ln [6x2 + 2]
Now differentiating f(x)
dy /dx (e(x+1) ln [6x2 + 2])
On applying chain rule:
3

f’(x) = e(x+1) ln [6x2 + 2] * d/dx ((x+1) ln [6x2 + 2] )
On solving d/dx ((x+1) ln [6x2 + 2] ) Using product rule:
6x (x+1) / (3x2 + 1) + ln (6x2 + 2)
Thus:
f’(x) = e(x+1) ln [6x2 + 2] * { 6x (x+1) / (3x2 + 1) + ln (6x2 + 2) }
e(x+1) ln [6x2 + 2] = (6x2 + 2)x+1
On substituting this value in above equation we have:
First derivative f ‘ (x) = (6x2 + 2)x+1 * { 6x (x+1) / (3x2 + 1) + ln (6x2 + 2) }
QUESTION 2
f(x) = (1 – [1 / 3] x) exp (-x2 + 5x - 6)
Solution
When limit x→∞
exp (-x2 + 5x - 6) = exp (-∞ ) = 0
Thus f (x) = 0
QUESTION 3
f(x) = 1 / [1 – (1+1/√3)x] exp (-x2 + (3 + √3 ) x – 3 √3 )
Domain = (1 / (1+1/√3), ∞)
Solution:
(a) Range
At x = ∞
f(x) = 1 / [1 – (1+1/√3) ∞] exp (-∞)
= [1/∞] * exp (-∞)
= 0
At x = ∞ f(x) = 0
Similarly :
1 / (1+1/√3) = 1.577
So at x = 1.577
4

f(x) = 1 / [1 – (1+1/√3) 1.57] exp (-(1.57)2 + (3 + √3 ) 1.57 – 3 √3 )
= -0.67 * exp (-2.46 +7.42 -5.19 )
= -0.67 * exp (-0.23 )
= - 0.52
At x = 1 / (1+1/√3) f(x) = -0.52 (negative value )
Thus range of function is (-0.52 , 0)
(b) Strictly decreasing interval
f(x) = 1 / [1 – (1+1/√3)x] exp (-x2 + (3 + √3 ) x – 3 √3 )
f(x) = uv
where u = 1 / [1 – (1+1/√3)x] and v = exp (-x2 + (3 + √3 ) x – 3 √3 )
f’(x) = uv’ +u’v
u’ = du /dx = {1 / [1 – (1+1/√3)x] }2 d/dx ([1 – (1+1/√3)x])
= {1 / [1 – (1+1/√3)x] }2 * [(-1-√3) /√3)]
u’ = [-√3 – 3 ] / [√3 – (√3 + 1) x] 2
Similarly
v = exp (-x2 + (3 + √3 ) x – 3 √3 )
v’ = dv /dx = exp (-x2 + (3 + √3 ) x – 3 √3 ) * d/dx [ (-x2 + (3 + √3 ) x – 3 √3 ) ]
d/dx [ (-x2 + (3 + √3 ) x – 3 √3 ) ] = (-2x +3 + √3)
v’ = (-2x +3 + √3) * exp (-x2 + (3 + √3 ) x – 3 √3 )
Thus first derivative is expressed as:
f’(x) = { 1 / [1 – (1+1/√3)x] * (-2x +3 + √3) * exp (-x2 + (3 + √3 ) x – 3 √3 ) } + {[-√3 – 3 ] / [√3
– (√3 + 1) x] 2 * exp (-x2 + (3 + √3 ) x – 3 √3 )}
f’(x) = exp (-x2 + (3 + √3 ) x – 3 √3 ) [{ 1 / [1 – (1+1/√3)x] * (-2x +3 + √3) + {[-√3 – 3 ] / [√3 –
(√3 + 1) x] 2]
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