Math 103. Homework 10. Solutions

1. Determine the intervals where the function

f(x)=x2 e-x

is increasing and the intervals where it is decreasing. Solution. The derivative

f¢(x) = 2x e-x + (-1) x2 e-x = (2x-x2)e-x

Setting f¢(x) = 0 gives 2x-x²=0 (because e^-x > 0 for all x). Therefore f¢(x) = 0 for x=0,2.

The sign diagram for the derivative is

so that f(x) is increasing on (0,2) and decreasing on (-¥,0)È(2,¥).

2. Determine the intervals of concavity and inflection points for the function

f(x) = x2 lnx

Solution. The domain of f(x) is (0,¥).

We compute f¢(x) = 2x lnx + x² (1/x) and the second derivative

f¢¢(x) = 2lnx + 2 +1 = 2lnx +3

Setting f¢(x)=0 gives

lnx = = 3 2

or

x = e-3/2

The sign diagram for the second derivative is

Therefore f(x) is concave down on (0,e^-3/2) and concave up on (e^-3/2,¥)

3. Use logarithmic differentiation to find an equation for the tangent line to the graph of

f(x) = xlnx

at the point (e,e).

Solution. We write

lnf(x) = ln(xlnx) = ( lnx) (lnx)

and taking derivatives

f¢(x) f(x) = 2lnx x

so that

f¢(x) = 2lnx x f(x) = 2lnx x xlnx

Thus

f¢(e) = 2lne e elne = 2(1) e e1 = 2

An equation for the line through (e,e) with slope 2 is

y = 2 x -e

4. Find the present value of $119,346 due in 4 yr at an interest rate of 10%/year compounded continuously.

Solution. The formula for continuous compounding of interest is

A=Pert

We know that A=119, 346, r=0.1 and t=4. Thus the present value is

P = (119,346)e(-0.1)(4) » 80,000

5. A hotel was purchased by a conglomerate for $4.5 million and sold 5 years later for $8.2 million. Find the annual rate of return, compounded continuously.

Solution. We use the formula

A=P ert

Here A = 8.2, P=4.5 and t=5. We have to find r. Plugging in these values into the formula we obtain

8.2 = 4.5 er 5

Hence

e5r = 8.2 4.5

and taking natural logarithms

5r = ln 8.2 4.5

or

r = 1 5 ln 8.2 4.5 = 0.120011

or 12%/year.