Question

find max and min value of function f(x)=7cosx-24sinx+5​

Answer 1

Answer:

12 & -19

Step-by-step explanation:

f(x) = 7 cos x - 24 sin x + 5

f '(x) = - 7 sin x -24 cos x

for f ' (x) = 0

- 7 sin x -24 cos x = 0

7 sinx = - 24 cos x

for f " (x) = - 7 cos x + 24 sin x

i,e. f " (x) = - 7 cos x + 24 ( - 24/7 cos x )

f " (x) = - 625/7 cos x

for x = 0° f " (x) = - 625/7 < 0 & f is max

for x = 90° f " (x) = 0 & f is max/min

for x = 180° f " (x) = 625/7 > 0 f is min

hence...

f(0°) = 7 cos 0° -24sin 0° + 5= 12 (max)

f(90°)= 7cos 90°-24sin90°+5 = -19 (min)

f(180°) =7cos 180° -24 sin180° +5 = -2(min)

therefore...

f max = 12 & f min = -19

Answer 2

$\bull \: \: \: \rm \underline \bold{Given :}$

$\sf \mapsto \: f(x) = \: 7 \cos \: x \: - 24 \: \sin \: x \: + \: 5$

$\bull \: \: \: \rm \underline \bold{Solution :}$

$\: \: \: \: \: \: \: \: \: \: \: \sf \: f(x) = \: 7 \cos \: x \: - 24 \: \sin \: x \: + \: 5$

$\sf \: \boxed{} \: f(x) \: will \: be \:max \: if \: -24 \: \sin \: x \: = \: 0$

$\sf hence,$

$\: \: \: \: \: \: \: \: \: \: \sf \: f(x)_{max}\:= 7 \cos \: x - 24 \sqrt{1 - { \cos}^{2}x } + 5$

$\sf \: Putting \: value \: of \: \cos \: x \: \\ \sf \: so \: that \: - 24 \sqrt{1 - { \cos}^{2}x } = 0$

$\implies \sf \: f(x) = 7 \times 1 - 24 \times 0 + 5$

$\implies \sf \: { \underline {\boxed {\sf{ f(x)_{max}= 12}}}}$

Now,

$\sf \: \boxed{} \: f(x) \: will \: be \:minimum \: if \: 7 \: \cos \: x \: = \: 0$

$\: \: \: \: \: \: \: \: \: \: \sf \: f(x)_{max}\:= 7 \cos \: x - 24 \sqrt{1 - { \cos}^{2}x } + 5$

$\sf hence,$

$\: \: \: \: \: \: \: \: \: \: \sf \: f(x)_{max}\:= 7 \sqrt{1 - { \sin}^{2}x } \: - 24 \sin \: x+ 5$

$\sf \: Putting \: value \: of \: \cos \: x \: \\ \sf \: so \: that \: woud\: 7 \sqrt{1 - { \sin}^{2}x } = 0$

$\implies \sf \: f(x)_{min}\:= 7 \sqrt{1 - 1 } \: - 24 \times 1+ 5$

$\implies \sf \: \: f(x)_{min}\: = - 24 + 5$

$\implies { \underline{\boxed{\sf \: {f(x)_{min}\: = - 19}}}}$

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