Find and classify all local extreme values of `h(x)=x(log_2 x)^2` .

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embizze | High School Teacher | (Level 2) Educator Emeritus

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Find and classify all local extreme values of `h(x)=x(log_2x)^2` .

**Assuming this is a real valued function **

(1) In order to find the extrema we take the first derivative. We use the product rule, the chain rule, and `d/(dx)(log_ax)=1/((lna)x)` where lna is the natural logarithm of a. Also `log_ax=1/(lna)lnx` . Thus:

`h'(x)=(log_2x)^2+x(2)(log_2x)(1/((ln2)x))`

`=((lnx)/(lna))^2+(2x)/(ln2)*1/(xln2)`

`=(lnx(lnx+2))/((ln2)^2)`

(2) The extrema occur at critical values for `h` , which include when `h'(x)=0` or when `h'(x)` fails to exist. `h'(x)` fails to exist in the reals if `x<=0` , but these values are not in the domain of `h` , otherwise `h'(x)` is continuous for all `x>0` .

`h'(x)=0` implies that `lnx(lnx+2)=0 => lnx=0` or `lnx=-2` .

If `lnx=0,x=1` and if `lnx=-2,x=e^(-2)` . Thus the only candidates for extrema are x=1 and `x=e^(-2)` .

(3) We apply the first derivative test. We check the value of `h'(x)` at points in the intervals `(0,e^(-2)),(e^(-2),1),(1,oo)` .

`h'(.1)>0,h'(.5)<0,h'(2)>0` so `x=e^(-2)` is a local maximum and x=1 is a local minimum.

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hala718 | High School Teacher | (Level 1) Educator Emeritus

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`h(x)= x(log_2 x)^2 `

`==gt h'(x)= (x)' (log_2 x)^2 + x [(log_2 x)^2]' `

`==gt g'(x)- (log_2 x)^2 + x(2log_2 x)(1/(xln2)) `

`==gt h'(x)= (log_2 x)^2 + (2xlog_2 x)/ln2 `

`==gt (log_2 x)^2 + (2log_2 x)/ln2 = 0 `

`==gt log_2 x (log_2 x + 2/ln2) =0 `

`==gt log_2 x = 0``==gt x = 2^0 = 1 `

`==gt log_2 x + 2/ln2 = 0 `

`==gt log_2 x = -2/ln2 ==gt x = 2^(-2/ln2) ==gt x = (1/2)^(2/ln2)`

`` ==> Then, the function has extremes values at the points x= 1 and `x= (1/2)^(2/ln2)`