Limits Determine a function f and a number a such that limit as h approaches zero ((2+h)^6)-64)/h =f'(a) .

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sciencesolve | Teacher | (Level 3) Educator Emeritus

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You need to remember limit definition of derivative such that:

`f'(a) = lim_(h-gt0) (f(h+a) - f(a))/h`

Comparing this definition,`lim_(h-gt0) (f(h+a) - f(a))/h` , to what problem provides, `lim_(h-gt0) ((h+2)^6-64)/h =f'(a),` yields `a=2`  and `f(2)=2^6 = 64` , thus `f(x) = x^6` .

Hence, derivative of the function `f(x)=x^6`  at `x=2`  is `f'(x)=6x^5`  such that:

`lim_(h-gt0) ((h+2)^6-2^6)/h = lim_(h-gt0) ((h+2-2)((h+2)^5 + 2(h+2)^4 + 4(h+2)^3 + 8(h+2)^2 + 16(h+2) + 32))/h`

Reducing h yields:

`lim_(h-gt0) ((h+2)^6-2^6)/h = lim_(h-gt0) ((h+2)^5 + 2(h+2)^4 + 4(h+2)^3 + 8(h+2)^2 + 16(h+2) + 32))`

You need to substitute 0 for h in equation under limit such that:

`lim_(h-gt0) ((h+2)^5 + 2(h+2)^4 + 4(h+2)^3 + 8(h+2)^2 + 16(h+2) + 32)) = 2^5 + 2*2^4 + 4*2^3 + 8*2^2 + 16*2 + 32`

`lim_(h-gt0) ((h+2)^5 + 2(h+2)^4 + 4(h+2)^3 + 8(h+2)^2 + 16(h+2) + 32)) = 32 + 32 + 32 + 32 + 32 + 32`

`lim_(h-gt0) ((h+2)^5 + 2(h+2)^4 + 4(h+2)^3 + 8(h+2)^2 + 16(h+2) + 32)) = 6*32 = 6*2^5`

`f'(2) = 6*2^5 = 192`

Hence, evaluating the function and "a" under given conditions yields `a = 2`  and `f(x) = x^6` . 

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