Question 1: f, g and H are functions such that
H(x) = (f g)(x)
and
f(1) = 36, f '(2) = 3, f '(1) = 4, g(1) = 9, g '(1) = 1.
Is the slope of the tangent line to the graph of H at x = 1 positive, negative or equal to zero?
Solution to Question 1:

G is the product of functions f and g hence.
G '(x) = f '(x) g(x) + f(x) g '(x)

Substitute x by 1 in the above to obtain.
G '(1) = f '(1) g(1) + f(1) g '(1) = (4)(9) + (36)(1) = 0

Since the value of the derivative of H at x = 1 is equal to the slope of the tangent line at x = 1, then the slope is also equal to zero and therefore the tangent line is parallel to the x axis.
Question 2: Function f is given by
f(x) = a x^{ 2} + b x + c
Find a, b and c so that f (0) = 3, f '(1) = 1 and f "(2) = 4, where f ' and f " are the first and second derivatives of function f.
Solution to Question 2:

We first use f (0) = 3 to write the following equation
f(0) = a (0)^{ 2} + b (0) + c = 3

Which gives
c = 3

We next calculate the first and second derivatives.
f '(x) = 2 a x + b
f "(x) = 2 a

We first use f "(2) = 4 to find a as follows
f "(2) = 2 a = 4

Solve for a to find
a = 2

We now use f '(x) = 2 a x + b and f '(1) = 1 to find b
f '(1) = 2 a (1) + b = 1

Substitute a by 2 and write an equation in b only
2 (2) (1) + b = 1

Solve for b to obtain
b = 3

The values of the 3 parameters a, b and c are
a = 2, b = 3 and c = 3.
Question 3: Let f be the function defined by f(x) = x^{3} + x and function g defined by g(x) = f^{ 1}(x) (inverse function). what is the value of g'(2)?
Solution to Question 3:

We first apply function f to both sides of g(x) = f^{ 1}(x) and write
f(g(x)) = f(f^{ 1}(x)) = x

We now have an equation of the form f(g(x)) = x which we differentiate (both sides), applying the chain rule to the left side, as follows
f '(g) . g'(x) = 1

We now set x = 2 in the above equation
f '(g(2)) . g '(2) = 1

We now calculate the first derivative of f as follows
f '(x) = 3 x ^{2} + 1

We now need the value of g(2). Note that f(1) = 2. Therefore g(2) = f^{ 1}(2) = 1. Hence
f '(1) . g '(2) = 1

We now calculate f '(1) using f '(x) = 3 x ^{2} + 1
f '(1) = 4

hence g '(2) = 1 / 4
Question 4: Let functions f, g and h be related as follows:
g(x) = f ^{ 1}(x), h(x) = (g(x))^{ 5}, f(6) = 10 and f '(6) = 12
Calculate h '(10).
Solution to Question 4:

We first calculate h '(x).
h '(x) = 5 g '(x) g(x)^{ 4}

Which gives
h '(10) = 5 g '(10) g(10)^{ 4}

g(10) is calculauted as follows
g(10) = f ^{ 1}(10) = 6

We now need to calculate g '(10). Apply function f to both sides of the relation g(x) = f ^{ 1}(x)
f (g(x)) = f(f ^{ 1}(x)) = x
Which gives f (g(x)) = x

Differentiate both sides of the equation obtained, using the chain rule to the term on the left.
f '(g) . g '(x) = 1

We now set x = 10 in the above equation
f '(g(10)) . g '(10) = 1

g(10) has already been calculated and is equal to 6, hence
f '(6) . g '(10) = 1

g '(10) is given by
g '(10) = 1 / f '(6) = 1 / 12

and h '(10) is given by
h '(10) = 5 (1/12) 6 ^{ 4} = 540
More references on calculus
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