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

1 Answer
Aug 28, 2016

f(35/4)f(354) is a local maximum of 37/4374

Explanation:

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

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]}