Question #653fb

2 Answers
Mr. Raj
Aug 17, 2017

The gradient of #f(x)=1/sqrtx# when #x=1# is #-1/2#

Explanation:

what you are looking for is the instantaneous gradient at #x=1#. To do so, you must find the derivative of the function, and substitute the #x=1# into the derivative to find the gradient.

First, we need to derive the function. To do that, simplify the function to make it easier to derive.
Hence:

#f(x)=1/sqrtx#

#=>f(x)=1/x^(1/2)#

#=>f(x)=x^(-1/2)#

Hence we derive this (remember, if #f(x)=x^n#, then the derivative is #f'(x)=nx^(n-1)#)
So:

#=>f'(x)=-1/2x^(-1/2-1)#

#=>f'(x)=-1/2x^(-3/2)#

Then simplify this (just to make it look better)

#=>f'(x)=-1/(2x^(3/2))#

#=>f'(x)=-1/(2xsqrtx)#

To find the gradient when #x=1#, substitute #x=1# into the equation above

#f'(1)=-1/(2(1)sqrt(1))#

#=>f'(1)=-1/(2)#

Hence the gradient of #f(x)=1/sqrtx# when #x=1# is #-1/2#

Jim H · mason m
Aug 17, 2017

To use limits, please see below.

Explanation:

The gradient at #x=1# can be found be evaluating

#lim_(hrarr0)(f(1+h)-f(1))/h# OR #lim_(xrarr1)(f(x) - f(1))/(x-1)#

Evaluation of these limits is quiet similar. Here is the second:

#lim_(xrarr1)(f(x) - f(1))/(x-1) = lim_(xrarr1) (1/sqrtx-1/sqrt1)/(x-1)#

#= lim_(xrarr1) ((1-sqrtx)/sqrtx)/((x-1)/1)#

#= lim_(xrarr1) ((1-sqrtx)/sqrtx * 1/(x-1))#

One Method

#= lim_(xrarr1) -(sqrtx-1)/(sqrtx (x-1))#

#= lim_(xrarr1) -(sqrtx-1)/(sqrtx (sqrtx-1)(sqrtx+1))#

#= lim_(xrarr1) -1/(sqrtx(sqrtx+1))#

# = -1/(1(1+1)) = -1/2#

Another method

from #" "# #" "# #= lim_(xrarr1) ((1-sqrtx)/sqrtx * 1/(x-1))#

#= lim_(xrarr1) (1-sqrtx)/(sqrtx(x-1))#

#= lim_(xrarr1) ((1-sqrtx)(1+sqrtx))/(sqrtx(x-1)(1+sqrtx))#

#= lim_(xrarr1) (1-x)/(sqrtx(x-1)(1+sqrtx))#

#= lim_(xrarr1) (-(x-1))/(sqrtx(x-1)(1+sqrtx))#

#= lim_(xrarr1) (-1)/(sqrtx(1+sqrtx))#

# = -1/(1(2)) = -1/2#

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