Calculus Help - BasilMarket Forum

Calculus Help

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thread starter satoraifu101 Feb 21 2010 + PM | QUOTE | PERMALINK | REPORT

Calculus AB HELP.! if the rate of change of f(x) = 2^x, at x =4 is twie as large as the rate of change at x=a, then a=? im sort of stuck on this problem... so wut i did was f ^' (x) = (ln2)2^x then im stuck.. can someone show me how to do this problem step by step with work >_>

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poke4554 02/21/10 + PM | QUOTE | PERMALINK | REPORT

omg i hate AB, so tricky and confusing but i got a = 3 what i did was f'x = 2^xln2 at x=4 which is 16ln2 and i divided that by half 8ln2 which is also 2^3ln2 i hope that helped

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DarkWater 02/21/10 + PM | QUOTE | PERMALINK | REPORT

Well, first find the rate of change at x=4. Since you have f', that's just substitution. It's 16ln2. Half of that is 8 ln 2, so you just set the derivative equal to that: 8 ln2 = (ln2) (2^x) 2^x = 8 x=3 a = 3

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poke4554 02/21/10 + PM | QUOTE | PERMALINK | REPORT

^^^^^^^^^^^^^^^^ Hi five!

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satoraifu101 02/21/10 + PM | QUOTE | PERMALINK | REPORT

how about if x = 5?

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poke4554 02/21/10 + PM | QUOTE | PERMALINK | REPORT

if x=5 then a=4

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MechaSoul 02/21/10 + PM | QUOTE | PERMALINK | REPORT

just think it through, the rate of change at any value x is ln(2)*2^x at x = 4 that's 16ln2 so set that equal to 2*ln(2)*2^a and solve for a (twice the rate of change at x = a)