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MAXIMA AND MINIMA Flipbook PDF
PART-II
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APPLICATION OF DERIVATIVES Manu
07/10/2020
Manu
APPLICATION OF DERIVATIVES
LOCAL MAXIMA AND LOCAL MINIMA
QUESTION07 Find the local maxima and local minima, if any,, if any, of the function f (x) = x 2 ANSWER07 f (x) = x 2 f 0 (x) = 2x 2x = 0 ⇒ x = 0 f ”(x) = 2, ∴ f ”(0) > 0 Hence f (x) has local minimum at x=0 and local minimum values is f(0)=0
Manu
APPLICATION OF DERIVATIVES
LOCAL MAXIMA AND LOCAL MINIMA QUESTION08 Find Find the local maxima and local minima, if any,, if any, of the function f (x) = x 3 − 3x ANSWER08 f (x) = x 3 − 3x f 0 (x) = 3x 2 − 3 = 3(x 2 − 1) f 0 (x) = 0 ⇒ x 2 − 1 = 0, ∴ x = ±1 f ”(x) = 6x at x = 1, f ”(1) = 6 > 0 Hence f (x) has local minimum at x=1 and local minimum values is f (1) = 13 − 3 = −2 at x = −1, f ”(−1) = −6 < 0 Hence f (x) has local maximum at x=-1 and local maximum values is f (−1) = (−1)3 − 3(−1) = 2 Manu
APPLICATION OF DERIVATIVES
LOCAL MAXIMA AND LOCAL MINIMA QUESTION09 Find the local maxima and local minima, if any,, if any, of the function f (x) = sin x + cos x, 0 < x < π2 ANSWER09 f (x) = sin x + cos x f 0 (x) = cos x − sin x f 0 (x) = 0 ⇒ cos x − sin x = 0, ∴ cos x = sin x x=
π 4
∈ (0, π2 )
f ”(x) = − sin x − cos x, ∴ f ”( π4 ) = − sin π4 − cos π4 =< 0 Hence f (x) is maximum at x =
π 4
Maximum value is f ( π4 ) = sin π4 + cos π4 √ = √12 + √12 = √22 = 2 Manu
APPLICATION OF DERIVATIVES
LOCAL MAXIMA AND LOCAL MINIMA QUESTION10 Find the local maxima and local minima, if any,, if any, of the function f (x) = x 3 –6x 2 + 9x + 15 ANSWER10 f (x) = x 3 –6x 2 + 9x + 15 f 0 (x) = 3x 2 − 12x + 9x f 0 (x) = 0 ⇒ 3(x 2 − 4x + 3) = 0, ∴ 3(x − 1)(x − 3) = 0 x = 1, 3 f ”(x) = 6x − 12, ∴ f ”(1) = 6 − 12 = −6 < 0 Hence f (x) is maximum at x = 1 Maximum value is f (1) = 13 − 6(1)2 + 9 + 15 = 19 f ”(3) = 6(3) − 12 = 6 > 0 Manu
APPLICATION OF DERIVATIVES
LOCAL MAXIMA AND LOCAL MINIMA
ANSWER 10...... Hence f (x) is minimmum at x = 3 Minimum value is f (3) = 33 − 6(3)2 + 9(3) + 15 = 15
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APPLICATION OF DERIVATIVES
QUESTION 11 Prove that function f (x) = e x does not have maxima or minima ANSWER 11 f (x) = e x f 0 (x) = e x f 0 (x) = 0 if e x = 0 But e x never gets value 0 for any real nuber x ∴ there does not exist any c ∈ R 3 f 0 (c) = 0 Hence f(x) does not have a maxima or minima
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APPLICATION OF DERIVATIVES
QUESTION 12 Prove that function f (x) = log x does not have maxima or minima ANSWER 12 f (x) = log x f 0 (x) =
1 x
log x is defined only for postive real numbers f 0 (x) > 0∀real numbers But
1 x
never gets value 0 for any real number x
∴ there does not exist any c ∈ R 3 f 0 (c) = 0 Hence f(x) does not have a maxima or minima
Manu
APPLICATION OF DERIVATIVES
QUESTION 13 Show that y = x 3 + x 2 + 2x + 1 does not attain the maximum or minimum value for any real number. ANSWER 13 f (x) = x 3 + x 2 + 2x + 1 f 0 (x) = 3x 2 + 2x + 2 f 0 (x) = 0 ⇒ 3x 2 + 2x + 1 = 0 solving above quadratic equation x = ∴ there does not exist any c ∈ R 3
√ −1± 2i 3
f 0 (c)
∈ /R
=0
Hence f(x) does not have a maxima or minima
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APPLICATION OF DERIVATIVES
QUESTION 14 Find the absolute maximum value and the absolute minimum value of the function f (x) = x 3 , x ∈ [–2, 2] ANSWER 14 f (x) = x 3 f 0 (x) = 3x 2 f 0 (x) = 0 ⇒ 3x 2 = 0 ∴, x = 0 Now we find f(-2),f(0) and f(2) f (−2) = (−2)3 = −8 f (0) = 0 f (2) = 23 = 8 So absolute minimum at x=-2,minimum value is -8 absolute maximum at x=2,maximum value is 8 Manu
APPLICATION OF DERIVATIVES
QUESTION 15 Find the absolute maximum value and the absolute minimum value of the function f (x) = sin x + cos x, x ∈ [0, π] ANSWER 15 f (x) = sin x + cos x f 0 (x) = cos x − sin x f 0 (x) = 0 ⇒ cos x − sin x = 0 ∴, cos x = sin x x=
π 4
f (0) = sin 0 + cos 0 = 1 f ( π4 ) = sin π4 + cos π4 =
√1 2
+
√1 2
=
√2 2
=
√
2
f (π) = sin π + cos π = 0 + −1 = −1 So absolute minimum at x = π,minimum value is -1 √ absolute maximum atx = π4 ,maximum value is 2 Manu
APPLICATION OF DERIVATIVES
QUESTION 16 Find both the maximum value and the minimum value of value of the function f (x) = 3x 4 –8x 3 + 12x 2 –48x + 25 on the interval [0, 3]. ANSWER 16 f (x) = 3x 4 –8x 3 + 12x 2 –48x + 25 f 0 (x) = 12x 3 − 24x 2 + 24x − 48 f 0 (x) = 0 ⇒ 12(x 3 − 2x 2 + 2x − 4) = 0 12[x 2 (x − 2) + 2(x − 2)] = 0 12(x − 2)(x 2 + 2) = 0 x =2 f (0) = 25 f (2) = 3(2)4 − 8(2)3 + 12(2)2 − 48(2) + 25 f (2) = 48 − 64 + 48 − 96 + 25 = −39 f (3) = 3(3)4 − 8(3)3 + 12(3)2 − 48(3) + 25 Manu
APPLICATION OF DERIVATIVES
ANSWER-16.... f (3) = 243 − 216 + 108 − 144 + 25 = 16 So minimum at x = 2,minimum value is -39 maximum at x = 0,maximum value is 25
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APPLICATION OF DERIVATIVES
QUESTION 17 Find the maximum value of 2x 3 –24x + 107 in the interval [1, 3]. Find themaximum value of the same function in [–3, –1]. ANSWER 17 f (x) = 2x 3 –24x + 107 f 0 (x) = 6x 2 − 24 = 6(x 2 − 4) f 0 (x) = 0 ⇒ 6(x − 2)(x + 2) = 0 = 0 x = 2, −2 f (1) = 2 − 24 + 107 = 85 f (2) = 2(2)3 − 24(2) + 107 = 16 − 48 + 107 = 75 f (3) = 2(3)3 − 24(3) + 107 = 54 − 72 + 107 = 89 So minimum at x = 2,minimum value is 75 maximum at x = 3,maximum value is 89 Manu
APPLICATION OF DERIVATIVES