Chapter 6 Application of Derivatives Ex 6.5 Solutions
Question - 1 : - Find the maximum and minimum values, if any, of thefollowing functions given by
(i) f(x) =(2x − 1)2 + 3 (ii) f(x) =9x2 +12x + 2
(iii) f(x) =−(x − 1)2 + 10 (iv) g(x)= x3 +1
Answer - 1 : -
(i) The given function is f(x) =(2x − 1)2 + 3.
It can be observed that (2x − 1)2 ≥ 0 for every x ∈ R.
Therefore, f(x) =(2x − 1)2 + 3 ≥ 3 forevery x ∈ R.
The minimum value of f isattained when 2x − 1 = 0.
2x − 1 = 0 ⇒ 
∴Minimum value of f =
= 3 Hence, function f doesnot have a maximum value.
(ii) The given function is f(x) =9x2 +12x + 2 = (3x + 2)2 −2.
It can be observed that (3x +2)2 ≥ 0 for every x ∈ R.
Therefore, f(x) =(3x + 2)2 − 2 ≥ −2 forevery x ∈ R.
The minimum value of f isattained when 3x + 2 = 0.
3x + 2 = 0 ⇒ 
∴Minimum value of f =
Hence, function f doesnot have a maximum value.
(iii) The given function is f(x) = − (x − 1)2 +10.
It can be observed that (x −1)2 ≥ 0 for every x ∈ R.
Therefore, f(x) =− (x − 1)2 + 10 ≤ 10 forevery x ∈ R.
The maximum value of f isattained when (x − 1) = 0.
(x − 1) = 0 ⇒ x = 1
∴Maximum value of f = f(1)= − (1 − 1)2 + 10 = 10
Hence, function f doesnot have a minimum value.
(iv) The given function is g(x)= x3 +1.
Hence, function g neitherhas a maximum value nor a minimum value.
Find the maximum and minimum values, if any, of thefollowing functions given by
(i) f(x) = |x +2| − 1 (ii) g(x) = − |x + 1| + 3
(iii) h(x) = sin(2x)+ 5 (iv) f(x) = |sin 4x + 3|
(v) h(x) = x +4, x
(−1, 1)
Answer - 2 : - (i) f(x)= 
We know that 
for every x ∈ R. Therefore, f(x) = 
for every x ∈ R.The minimum value of f is attainedwhen 
.
∴Minimum value of f = f(−2)
Hence, function f does nothave a maximum value.
(ii) g(x) =
We know that 
for every x ∈ R. Therefore, g(x) = 
for every x ∈ R. The maximum value of g is attainedwhen
∴Maximum value of g = g(−1)= 
Hence, function g does nothave a minimum value.
(iii) h(x) = sin2x +5
We know that − 1 ≤ sin 2x ≤ 1.
⇒ − 1 + 5 ≤ sin 2x + 5 ≤ 1 + 5
⇒ 4 ≤ sin 2x + 5 ≤ 6
Hence, the maximum and minimum valuesof h are 6 and 4 respectively.
(iv) f(x) =
We know that −1 ≤ sin 4x ≤ 1.
⇒ 2 ≤ sin 4x + 3 ≤ 4
⇒ 2 ≤≤ 4 Hence, the maximum and minimum valuesof f are 4 and 2 respectively.
(v) h(x) = x +1, x ∈ (−1, 1)
Here, if a point x0 isclosest to −1, then we find 
for all x0 ∈ (−1, 1). Also, if x1 isclosest to 1, then 
for all x1 ∈ (−1, 1). Hence, function h(x) hasneither maximum nor minimum value in (−1, 1).
Question - 3 : - Find the local maxima and local minima, if any, of thefollowing functions. Find also the local maximum and the local minimum values,as the case may be:
(i). f(x)= x2 (ii). g(x)= x3 −3x (iii). h(x) =sinx +cosx, 0<
(iv). f(x) =sinx − cos x, 0 < x <2π
(v). f(x) = x3 −6x2 +9x +15
(vi). 
(vii). 
(viii). 
Answer - 3 : -
(i) f(x)= x2
Thus, x =0 is the only critical point which could possibly be the point of local maximaor local minima of f.
We have
, which is positive.
Therefore, by second derivative test, x =0 is a point of local minima and local minimum value of f at x =0 is f(0) = 0.
(ii) g(x)= x3 −3x
By second derivative test, x = 1 is a point of local minima and local minimum valueof g at x = 1 is g(1)= 13 − 3 = 1 − 3 = −2. However,
x = −1 is a point of local maxima and local maximumvalue of g at
x = −1 is g(1) = (−1)3 −3 (− 1) = − 1 + 3 = 2.
(iii) h(x) =sinx +cosx, 0< x <
Therefore, by secondderivative test,
is a point of localmaxima and the local maximum value of h at is 
Therefore, by secondderivative test,
is a point of localmaxima and the local maximum value of f at
is
However,
is a point of localminima and the local minimum value of f atis 
(v) f(x)= x3 −6x2 +9x + 15
Therefore, by second derivative test, x =1 is a point of local maxima and the local maximum value of f at x = 1 is f(1) = 1 − 6 + 9 + 15 =19. However, x = 3 is a point of local minima and the local minimumvalue of f at x = 3 is f(3) = 27 − 54 + 27 + 15 = 15.
(vi) 
Since x >0, we take x = 2.
Therefore, by secondderivative test, x = 2 is a point of local minima and the localminimum value of g at x =2 is g(2) =
(vii) 
Now, for values closeto x = 0 and to the left of 0,
Also, for values closeto x = 0 and to the right of 0,
Therefore, by firstderivative test, x = 0 is a point of local maxima and the localmaximum value of
(viii) 
Therefore, by secondderivative test,
is a point of local maximaand the local maximum value of f at is
Question - 4 : - Prove that the following functions do not have maxima orminima:
(i) f(x) = ex (ii) g(x)= logx
(iii) h(x) = x3 + x2 + x +1
Answer - 4 : -
(i) We have,
f(x)= ex
Now, if 
. But, the exponential function can neverassume 0 for any value of x.
Therefore, there does notexist c∈ R such that
Hence, function f does nothave maxima or minima.
(ii) We have,
g(x)= log x
Therefore, there does notexist c∈ R such that
.
Hence, function g does nothave maxima or minima.
(iii) We have,
h(x)= x3 + x2 + x +1
Now,
h(x)= 0 ⇒ 3x2 +2x + 1 = 0 ⇒ 
Therefore, there does notexist c∈ R such that
.
Hence, function h does nothave maxima or minima.
Question - 5 : - Find the absolute maximum value and the absolute minimumvalue of the following functions in the given intervals:
(i) 
(ii) 
(iii) 
(iv) 
Answer - 5 : -
(i) The given function is f(x)= x3.
Then, we evaluate the value of f atcritical point x = 0 and at end points of the interval [−2,2].
f(0) = 0
f(−2) = (−2)3 =−8
f(2) = (2)3 =8
Hence, we can conclude that the absolutemaximum value of f on [−2, 2] is 8 occurring at x =2. Also, the absolute minimum value of f on [−2, 2] is −8occurring at x = −2.
(ii) The given function is f(x)= sin x + cos x.
Then, we evaluate the valueof f at critical point and at the end points of the interval[0, π].
Hence, we can conclude thatthe absolute maximum value of f on [0, π] is
occurring atand the absolute minimum value of f on[0, π] is −1 occurring at x = π.
(iii) The given function is
Then, we evaluate the valueof f at critical point x = 4 and at the endpoints of the interval
.
Hence, we can conclude thatthe absolute maximum value of f onis 8 occurring at x =4 and the absolute minimum value of f on is −10 occurring at x = −2. (iv) The given function is
.
Now,
2(x − 1) = 0 ⇒ x = 1
Then, we evaluate the value of f atcritical point x = 1 and at the end points of the interval[−3, 1].
Hence, we can conclude that the absolutemaximum value of f on [−3, 1] is 19 occurring at x =−3 and the minimum value of f on [−3, 1] is 3 occurringat x = 1.
Find the maximum profit that a company can make, if theprofit function is given by
p(x)= 41 − 72x − 18x2
Answer - 6 : - The profit function isgiven as p(x) = 41 − 72x − 18x2.
By secondderivative test, x=-2 is the point oflocal maxima of p.
Hence, the maximum profit that the company can make is113 units.
Find both the maximum value and the minimum value of
3x4 − 8x3 +12x2 − 48x + 25 on the interval [0, 3]
Answer - 7 : -
Let f(x) = 3x4 −8x3 + 12x2 − 48x +25.
Now,gives x = 2 or x2+ 2= 0 for which there are no real roots.
Therefore, we consider only x =2 ∈[0, 3].
Now, we evaluate the value of f atcritical point x = 2 and at the end points of the interval [0,3].
Hence, we can conclude that the absolutemaximum value of f on [0, 3] is 25 occurring at x =0 and the absolute minimum value of f at [0, 3] is − 39occurring at x = 2.
Question - 8 : - At what points in the interval [0, 2π], doesthe function sin 2x attain its maximum value?
Answer - 8 : -
Let f(x) = sin 2x.
Then, we evaluate thevalues of f at critical points
and at the end points of the interval [0,2π].
Hence, we can conclude thatthe absolute maximum value of f on [0, 2π] is occurring atand
Question - 9 : - What is the maximum value of the functionsin x + cos x?
Answer - 9 : -
Let f(x) = sin x +cos x.
Now, 
will be negative when(sin x + cos x) is positive i.e., when sin x andcos x are both positive. Also, we know that sin x andcos x both are positive in the first quadrant. Then, will be negative when
.
Thus, we consider.
∴By second derivative test, f will bethe maximum atand the maximum valueof f is
.
Question - 10 : - Find the maximum value of 2x3 −24x + 107 in the interval [1, 3]. Find the maximum value of thesame function in [−3, −1].
Answer - 10 : -
Let f(x) = 2x3 −24x + 107.
We first consider the interval [1, 3].
Then, we evaluate the value of f atthe critical point x = 2 ∈ [1, 3] andat the end points of the interval [1, 3].
f(2) = 2(8)− 24(2) + 107 = 16 − 48 + 107 = 75
f(1) = 2(1)− 24(1) + 107 = 2 − 24 + 107 = 85
f(3) = 2(27)− 24(3) + 107 = 54 − 72 + 107 = 89
Hence, the absolute maximum value of f(x)in the interval [1, 3] is 89 occurring at x = 3.
Next, we consider the interval [−3, −1].
Evaluate the value of f atthe critical point x = −2 ∈ [−3, −1]and at the end points of the interval [1, 3].
f(−3) = 2(−27) − 24(−3) + 107 = −54 + 72 + 107 = 125
f(−1) =2(−1) − 24 (−1) + 107 = −2 + 24 + 107 = 129
f(−2) =2(−8) − 24 (−2) + 107 = −16 + 48 + 107 = 139
Hence, the absolute maximum value of f(x)in the interval [−3, −1] is 139 occurring at x = −2.