6. let f and g be differentiable functions
with the following properties:
i. g(x) > 0 for all x
ii. f(0) = 1
if h(x) = f(x)g(x) and h'(x)
= f(x)g'(x), then what is f(x)?
(a) 0
(b) 1
(c) f'(x)
(d) e^x

With the given properties in the question, the option that represents the correct value of f(x) is;

Option B: 1

We are told that;

h(x) = f(x)g(x)

Thus;

h'(x) = (d/dx)[f(x)g(x)]

>> h'(x) = f(x)•g'(x) + g(x)•f'(x)

Now, we are given that;

h'(x) = f(x)g'(x)

Applying this property to our derivatives h'(x), we have;

f(x)g'(x) = f(x)•g'(x) + g(x)•f'(x)

Simplifying, f(x)g'(x) will cancel out to give us;

g(x)•f'(x) = 0

       Thus; f'(x) = 0

f(x) = d

Thus; f(0) = d

We are told that; f(0) = 1

By transitive property earlier proved, we can deduce that;

d = 1

Thus; f(x) = 1

Read more at; link

(b) 1

Step-by-step explanation:

To differentiate we will need the product rule:

.

We have , so the following equation is true by the transitive property:

By subtraction property we have:

Since , then we can divide both sides by :

This implies is constant.

So we have that where is a real number.

Since and , then by transitive property .

So .

Checking:

So the following conditions were met.

about 0.75 hours

step-by-step explanation:

not sure but it may be b, as i said i'm not sure.