Question

The function $f(x) = {e^{3x}}$ is strictly increasing function on $R$ ?

Answer 1

Hint : A function $f$ is said to be strictly increasing for every value of x the value of y will goes on increasing. The strictly increasing function is defined as if ${x_1}$ and ${x_2}$ belongs to $R$ we have ${x_1} < {x_2}$ then $f({x_1}) < f({x_2})$ .

Complete step-by-step answer :
Here in this question we have to find whether the given function is strictly increasing or not. As we know the definition of strictly increasing function that is if ${x_1} < {x_2}$ then $f({x_1}) < f({x_2})$ for all ${x_1},{x_2} \in R$ .
Given $f(x) = {e^{3x}}$
Suppose if we have ${x_1} < {x_2}$
Multiply 3 to above inequality we have
$\Rightarrow 3{x_1} < 3{x_2}$
Applying the exponent, we have
$\Rightarrow {e^{3{x_1}}} < {e^{3{x_2}}}$
We have $f(x) = {e^{3x}}$ therefore we can write as
$\Rightarrow f({x_1}) < f({x_2})$
We have proved the given condition that is ${x_1} < {x_2}$ then $f({x_1}) < f({x_2})$ for all ${x_1},{x_2} \in R$ . This proves that the given function $f(x) = {e^{3x}}$ is strictly increasing function on $R$ .

Another method :
Suppose we have to check the given function is strictly increasing, we have another method that is we have to differentiate the given function w.r.t, x and that function should be greater than zero that is $f'(x) > 0$ .
Now consider the given function $f(x) = {e^{3x}}$
Differentiate the above function w.r.t, x we get
$\Rightarrow f'(x) = \dfrac{d}{{dx}}({e^{3x}})$
We know that $\dfrac{d}{{dx}}({e^{ax}}) = a.{e^{ax}}$
By applying that we have
$\Rightarrow f'(x) = 3{e^{3x}}$
By differentiating we got the real positive number and that is greater than zero. Hence it satisfies the given condition that is $f'(x) > 0$ .
Therefore, the given function is strictly increasing function on $R$ .
By using two methods we have proved that the given function is strictly increasing function
Hence the function $f(x) = {e^{3x}}$ is strictly increasing function on $R$ .
Formula used:
1. If ${x_1}$ and ${x_2}$ belongs to $R$ we have ${x_1} < {x_2}$ then $f({x_1}) < f({x_2})$ for all ${x_1},{x_2} \in R$ .
2. $f'(x) > 0$ . $x \in R$ .

Note : Candidate can follow any one of the methods to check whether the function is strictly increasing function or not. We have to apply and check the condition and if it satisfies then function is strictly increasing and by the second method if we differentiate if we get a real positive number then function is strictly increasing.