Page 104 - How To Solve Word Problems In Calculus

Step4
For ease in differentiation, we rewrite t(x) using
exponents:

1 1
2
t(x) = (x + 1) 1/2 + (3 − x)
3 5
1 1
2 −1/2
t (x) = (x + 1) (2x) + (−1)
6 5
x 1
= √ −
2
3 x + 1 5
We set the derivative to 0 and solve for x.

x 1
0 = √ −
2
3 x + 1 5
1 x
= √
5 3 x + 1
2

2
5x = 3 x + 1
2
2
25x = 9(x + 1)
2
2
25x = 9x + 9
2
16x = 9
9
2
x =
16
3
x =
4

Step5
Before we jump to any conclusion about the route Frank
should take, let us examine the endpoints of the interval.
√
2
x + 1 3 − x
Recall that t(x) = + .
3 5
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