Problem 1: Calculate or estimate the following limits:

a) 

Multiply top and bottom by 2, then pull out the 2 on the top to get = 2 * 1 = 2.

b) 

The limit is the derivative of f(z) = 1/z at z = 3, which should be -1/3^2 = -1/9. You can also get a common denominator on the top: (3-z)/(3z) = -1*(z-3)/(3z). Dividing by (z-3) gives -1/(3z) which goes to -1/9, as z approaches 3.

c) 

Numerical estimates (x = 1.999 and 2.00003 say) are close to -9.885. This limit is also the negative of the derivative of 3^x at x = 2, which is -ln(3)*3^2.

d) 

The exponential will eventually overpower the polynomial, so the limit will be negative infinity.

e) 

At x = 5, the denominator is zero. A graph near x = 5 indicates a vertical asymptote at x = 5, with the function decreasing to negative infinity to the left of x = 5.

> plot((x-2)/(x^2-3*x-10),x=4..6,y=-10..10,discont=true,color=black);

Problem 2: Find the tangent line equation at x = -1, for f(x) = -x^2 + 2^x/ln(2).

The slope is f '(-1), and (1, f(1) ) is on the line.

> deriv:=diff(-x^2+2^x/ln(2),x);
slope := subs({x=-1},deriv);

> d:=subs({x=-1},-x^2+2^x/ln(2));
decimal_version:= evalf(d);

We therefore get y = 2.5(x + 1) -0.27865, for the tangent line equation.

Problem 3: If f(x) = e^(-x^2), is f(x) concave up or concave down at x = 1?

You can plot the derivative f '(x) near x = 1, and use the fact that f '(x) increasing is equivalent to f(x) concave up, while f '(x) decreasing means f(x) is concave down, OR check whether f ''(1) is positive (f(x) is concave up) or f ''(1) is negative (f(x) is concave down). Here are both tests.

> deriv1:=diff(exp(-x^2),x);
deriv2:=diff(deriv1,x);

> plot(deriv1,x=0..2,title=`f'(x) is inc at x = 1`);

> subs({x=1}, deriv2);
as_a_decimal:=evalf(%);

Both tests say f(x) is concave up at x = 1, take a look, its hard to tell from a graph of f(x):

> plot(exp(-x^2),x=0..2,title=`f(x) near x = 1`);

Problem 4: The table below gives the number of rabbits, R, at various times, t, measured in months, along with the growth rate of the rabbits at 0 and 1.6 months.

a) Estimate the growth rates at the remaining times.

b) Estimate a formula for R(t).

c) Estimate a formula (based on the table data) for R'(t).

d) Differentiate your formula for R(t) from part (b) and compare to your answer for part (c).

Using, for example (R(0.8) - R(0) ) / (0.8 -0) = (39 - 25)/0.8 as an estimate for the growth rate at t = 0.4 months, we get:

Here are the ratios (the slopes are not constant) from the rabbit column:

> ratios:=31/25.,39/31.,48/39.,60./48;

It looks like they are all about 1.245, so we should be able to find an exponential formula, R(t) = k*b^t, and these ratios should all be equal to b^0.4, so here is an estimate for b:

> b_value:=1.245^(1/0.4);

Since k = R(0) = 25, we can guess R(t) = 25 *1.73^t. This yields the following values for the number of rabbits at the times in the table:

> R_values:=map(t->25*1.73^t,[0.0,0.4,0.8,1.2,1.6]);

Those look good. Now, repeat this process for the R'(t) values:

> ratios2:=21.25/17.5, 26.25/21.25;

> b2val:=1.23^(1/0.4);

This gives R'(t) = k*1.68^t. Substituting in t = 0.4 and R' = 17.5, gives 17.5 = k*1.68^0.4, or

> k = solve(17.5=k*1.68^0.4,k);

> Rprime_values:=map(t->14.2*1.68^t,[0.4,0.8,1.2]);

Again, our formula values are close to the table data. Here is the derivative of our formula for R(t):

> 25*ln(1.73)*1.73^t;

Moderately close to 14.2*1.68^t.

Problem 5: Draw a graph of y = f(x), given the following information about its derivative:

a) For x < -2, f '(x) > 0, and the derivative is increasing. (so f(x) is increasing and concave up.)

b) For -2 < x < 1, f '(x) > 0 and the derivative is decreasing. (so f(x) is increasing and concave down.)

c) At x = 1, f '(x) = 0. (x = 1 is a hilltop on the graph of f(x).)

d) For x > 1, f '(x) < 0 and the derivative is decreasing. (so f(x) is decreasing and concave down.)

Part 2 of Math 170 exam 2 practice answers.

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