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The function f(x) = ax, 0 < a < 1 is
- a)increasing
- b)strictly decreasing on R
- c)neither increasing or decreasing
- d)decreasing
Correct answer is option 'D'. Can you explain this answer?
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Most Upvoted Answer
The function f(x) = ax, 0 < a < 1 isa)increasingb)strictl...
f(x) = ax
Taking log bth the sides, log f(x) = xloga
f’(x)/ax = loga
f’(x) = ax loga {ax > 0 for all x implies R,
for loga e<a<1 that implies loga < 0}
Therefore, f’(x) < 0, for all x implies R
f(x) is a decreasing function.
Taking log bth the sides, log f(x) = xloga
f’(x)/ax = loga
f’(x) = ax loga {ax > 0 for all x implies R,
for loga e<a<1 that implies loga < 0}
Therefore, f’(x) < 0, for all x implies R
f(x) is a decreasing function.
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Community Answer
The function f(x) = ax, 0 < a < 1 isa)increasingb)strictl...
f(x) = ax
Taking log bth the sides, log f(x) = xloga
f’(x)/ax = loga
f’(x) = ax loga {ax > 0 for all x implies R,
for loga e<a<1 that implies loga < 0}
Therefore, f’(x) < 0, for all x implies R
f(x) is a decreasing function.
Taking log bth the sides, log f(x) = xloga
f’(x)/ax = loga
f’(x) = ax loga {ax > 0 for all x implies R,
for loga e<a<1 that implies loga < 0}
Therefore, f’(x) < 0, for all x implies R
f(x) is a decreasing function.
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The function f(x) = ax, 0 < a < 1 isa)increasingb)strictly decreasing on Rc)neither increasing or decreasingd)decreasingCorrect answer is option 'D'. Can you explain this answer?
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