If - f-x- - x-5 - 20x-3 - 240x- then f-x- is nbsp-monotonic increasing everywheremonotonic decreasing only in - -0- -monotonic decreasing everywheremonotonic increasing only in - - - 0-

The correct option is A monotonic increasing everywheref(n)= x^5 20x^3+240x #160; f(n)=5x^4 60x^2+240 f'(n)=5x^4 60x^2+240 #160; #160; ...

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Byju's Answer

Standard XII

Mathematics

Condition for Monotonically Decreasing Function

If f(x) ...

Question

If $f (x) = x^{5} - 20 x^{3} + 240 x$ , then f(x) is

monotonic increasing everywhere

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monotonic decreasing only in

(0, \infty)

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monotonic decreasing everywhere

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monotonic increasing only in

(- \infty, 0)

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Solution

The correct option is A monotonic increasing everywhere
f(n)= $x^{5} - 20 x^{3} + 240 x$ $f (n) = 5 x^{4} - 60 x^{2} + 240$ $f^{'} (n) = 5 x^{4} - 60 x^{2} + 240$
$5 x^{2} - 60 x^{2} + 240 = 0$
$5 x^{4} - 60 x^{2} + 240 = 0$
$x^{2} - 12 x^{2} + 48 = 0$
D<0
So, f(n) is always increasing

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If $f (x) = x^{5} - 20 x^{3} + 240 x$ , then f(x) is

The correct option is A monotonic increasing everywhere
f(n)= $x^{5} - 20 x^{3} + 240 x$ $f (n) = 5 x^{4} - 60 x^{2} + 240$ $f^{'} (n) = 5 x^{4} - 60 x^{2} + 240$
$5 x^{2} - 60 x^{2} + 240 = 0$
$5 x^{4} - 60 x^{2} + 240 = 0$
$x^{2} - 12 x^{2} + 48 = 0$
D<0
So, f(n) is always increasing

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If f(x)=x5−20x3+240x, then f(x) is

The correct option is A monotonic increasing everywheref(n)=x5−20x3+240x f(n)=5x4−60x2+240 f′(n)=5x4−60x2+240 5x2−60x2+240=0 5x4−60x2+240=0 x2−12x2+48=0 D<0So, f(n) is always increasing

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FREE TEXTBOOK SOLUTIONS

STATE BOARDS

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If $f (x) = x^{5} - 20 x^{3} + 240 x$ , then f(x) is

The correct option is A monotonic increasing everywhere
f(n)= $x^{5} - 20 x^{3} + 240 x$ $f (n) = 5 x^{4} - 60 x^{2} + 240$ $f^{'} (n) = 5 x^{4} - 60 x^{2} + 240$
$5 x^{2} - 60 x^{2} + 240 = 0$
$5 x^{4} - 60 x^{2} + 240 = 0$
$x^{2} - 12 x^{2} + 48 = 0$
D<0
So, f(n) is always increasing