Question

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​

Answer 1

Answer:

(b) 1

Step-by-step explanation:

To differentiate $h(x)=f(x)g(x)$ we will need the product rule:

$h'(x)=f'(x)g(x)+f(x)g'(x)$.

We have $h'(x)=f(x)g'(x)$, so the following equation is true by the transitive property:

$f'(x)g(x)+f(x)g'(x)=f(x)g'(x)$

By subtraction property we have:

$f'(x)g(x)=0$

Since $g(x) \neq 0$, then we can divide both sides by $g(x)$:

$f'(x)=\frac{0}{g(x)}$

$f'(x)=0$

This implies $f(x)$ is constant.

So we have that $f(x)=c$ where $c$ is a real number.

Since $f(0)=1$ and $f(0)=c$, then by transitive property $1=c$.

So $f(x)=1$.

Checking:

$h(x)=1 \cdot g(x)$

$h'(x)=g'(x)=1 \cdot g'(x)$

So the following conditions were met.

Answer 2

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)

• Now, to differentiate this, we will apply product rule of differentiation to get the derivative of h(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)

• According to transitive property in maths, if P = Q and Q = S, Then we can say that; P = S

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

• We are told that g(x) > 0, thus, it means for g(x)•f'(x) to be equal to 0, then f'(x) must be equal to zero.

       Thus; f'(x) = 0

• When we differentiate an expression and we get zero, it means that expression initially was a constant number. This means that if that number is d, then;

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

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