How do you find local maximum value of f using the first and second derivative tests: #f(x) = x + sqrt(9 − x) #?

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Answer 1 · 2016-08-28T10:49:34

#f(35/4)# is a local maximum of #37/4#

#f(x) = x+sqrt(9-x)#

#f'(x) = 1+1/2(9-x)^(-1/2)* (-1) = 1-1/(2sqrt(9-x))#

For local maxima or minima #f'(x) =0#

#f'(x) =0 -> 1-1/(2sqrt(9-x)) =0#

#1/sqrt(9-x) =2#

#sqrt(9-x)=1/2#

#9-x =1/4#

#x=9-1/4 = 35/4#
#-> f(x) = 35/4 +sqrt(9-35/4)#

#= 35/4 + sqrt(1/4)#

#= 35/4 +1/2 = 37/4#

Hence #f(x)# has a turning point at #(35/4,37/4)#

Now consider #f''(x) = 0- d/dx1/2 (9-x)^(-1/2)#

#= -(-1/4(9-x)^(-3/2)*(-1))#

#= -1/(4(9-x)^(3/2))#

#f''(35/4) = -1/(4(9-35/4)^(3/2)) = -1/(4*(1/4)^(3/2))#

Hence #f''(35/4) < 0 -> f(35/4)# is a local maximum = #37/4#

This may be seen by the graph of #f(x)# in the region of #f(35/4)#:

graph{x+sqrt(9-x) [7.9024, 9.588, 8.735, 9.5774]}