So `100 = e^A = e^(ln(Z/10)) = Z/10` because the exponential function is the inverse of the natural logarithm function. Solving for `Z` gives `Z = 1000`.

For the second integral, `int_0^2 1/2 1/(u^2+1) du = {:1/2 tan^(-1)(u)|_0^2 = tan^(-1)(2) - tan^(-1)(0) = tan^(-1)(2)`
because `tan^(-1)(0) = 0`. Combining results, one has the answer
` ln(5)/8 + tan^(-1)(2)`.

(10 points) If `f(x) = x^2 + x + 2` for `x > 0`, what is the equation of the tangent line to the graph of `f^(-1)(x)` at the point `(4, 1)`? SHOW ALL WORK!

To find the equation of a line one needs the slope and a point. The point
is `(4,1)`. The slope is given by the derivative

where we used the formula for the derivative of the inverse function.
We know that there actually is an inverse function because `f'(x) = 2x + 1 > 0` for `x > 0` and so `f` is an increasing function and all increasing functions are one-to-one.

Using the point slope form of the equation of the line:

The function is `y = (2x + 1)/(3x + 5)`. Swap the variable names to get
`x = (2y+1)/(3y+5)` and solve for `y` in terms of `x`: `3xy + 5x = 2y + 1`,
`3xy - 2y = 1 - 5x`, `(3x-2)y = 1- 5x`, and so `y = (1-5x)/(3x-2)`.
The answer is `f^(-1)(x) = (1 - 5x)/(3x-2)`.

(5 points For your function `f^(-1)(x)` in part (a), show `f^(-1)(f(x)) = x`.

Substitute the expression for `f(x)` into that for `f^(-1)(x)` and simplify:

`f^(-1)(f(x))

= (1 - 5((2x+1)/(3x+5)))/(3((2x+1)/(3x+5))-2)`

`=((3x+5) - 5(2x+1))/(3(2x+1)-2(3x+5))`

`=(-7x)/(-7) = x`

as was to be shown.

SHOW ALL WORK on the following. Suppose `g(x)` is a function defined for all `x` and such that

`g(7) = 2`

`g'(7) = -5`

Let `F(x) = 3^(g(x))`.

(2 points) `F(7) = 3^(g(7)) = 3^2 = 9`

(2 points) `F'(7) = -45 ln(3)`

One calculates the derivative `F'(7) = (3^(g(x))' = ln(3) 3^(g(x)) g'(x)` using the chain rule. Substituting `x=7` gives `F'(7) = ln(3) 3^(g(7)) g'(7) = ln(3)3^2 (-5) = -45 ln(3)`.

(3 points) The equation for the tangent line to the graph of `F(x)` at `x=7` is `-45ln(3)(x - 7) + 9`

(3 points) The equation for the tangent line to the graph of `F^(-1)(x)` at `(F(7), 7)` is `x = -45ln(3)(y-7) + 9`.

The tangent line to the inverse is the reflection of the original tangent line
across the 45 degree line `y = x`. So, to obtain its equation, just swap the
variables `x` and `y`. If you want, you can also solve for `y` giving
`y = -(x-9)/(45 ln(3)) + 7`.

SHOW ALL WORK on the following.
(10 points) An observer is at ground level 100 feet from the base of a tower 300 feet high. She watches as an exterior elevator ascends from ground level. At time `t` seconds the elevator is at `4t` feet above the ground.

If `T(t)` is the angle (in radians) formed by her line of sight and the horizontal at time `t` seconds, then:

`T(t) = tan^(-1) ((4t)/100)` because `tan(T(t)) = (4t)/100`.

`T'(t) = 25/(625+t^2)

Differentiate using the chain rule:

`T'(t) = 1/(1+ (t/25)^2) 1/25 = 25/(625+t^2)`

`T'(7) = 25/(625 + 7^2) = 25/674`

SHOW ALL WORK on the following.(5 points) `lim_(x->oo) (4e^(2x)-5x^2+11)/(5e^(2x)-4x^2+11) =4/5`

The limit of the numerator and denominator is infinity and so one could
try to apply L'Hospital's rule repeatedly to get the answer. However, it
is quicker to rewrite the expression by dividing numerator and denominator
by `e^(2x)`. This gives