Question

The range of f(x) = e2x + 2ex + 3 is

Answer 1

The range of f(x) = e²ˣ + 2eˣ + 3 is (3, ∞)

Step-by-step explanation:

Given:

Function $f(x)=e^{2x}+e^x+3$

To find out:

The range of the f(x)

Solution:

$f(x)=e^{2x}+e^x+3$

$\implies f(x)=(e^x)^2+e^x+3$

If we take $e^x=t$

Then

A new function g(t) can be written as

$g(t)=t^2+2t+3$

Which is a quadratic polynomial

Here $b^2-4ac=2^2-4\times 1\times 3=-8<0$

Thus, the polynomial g(t) does not have real zeroes

Therefore, the polynomial g(t) will not cut the x-axis

And since a = 1 > 0

Therefore, this will lie above the x-axis

$g(t)= t^2+2t+3$

$\implies g(t)=t^2+2t+1+2$

$\implies g(t)=(t+1)^2+2$

Thus, the minimum value of g(t) will be when t = -1

The minimum value = 2

But the domain of g(t) will be the range of $t=e^x$

We know that range of $e^x$ is (0, ∞)

And at t = 0

$g(0)=3$

Thus, the range of f(x) will be from (3, ∞)

Hope this answer is helpful.

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