Q. \( y(x) = \frac{3}{2},\mathrm{e}^{2x} – \frac{1}{2} – x. \)

Problem

Analyze the function:

\(y(x) = \frac{3}{2}e^{2x} – \frac{1}{2} – x\)

Step 1: Identify the type of function

The function is:

\(y(x) = \frac{3}{2}e^{2x} – \frac{1}{2} – x\)

This function contains two main parts:

An exponential part:

\(\frac{3}{2}e^{2x}\)

and a linear part:

\(-x – \frac{1}{2}\)

So this is an exponential function combined with a linear expression.

Step 2: Understand each term separately

The first term is:

\(\frac{3}{2}e^{2x}\)

Here, \(e\) is Euler’s number, which is approximately:

\(e \approx 2.718\)

The expression \(e^{2x}\) means the exponent is \(2x\). As \(x\) increases, \(e^{2x}\) grows very quickly.

The coefficient \(\frac{3}{2}\) multiplies the exponential expression, so it stretches the exponential part vertically.

The second term is:

\(-\frac{1}{2}\)

This shifts the graph downward by \(\frac{1}{2}\).

The third term is:

\(-x\)

This is a linear term. It decreases as \(x\) increases.

Step 3: Find the domain

The domain is the set of all possible input values of \(x\).

The function is:

\(y(x) = \frac{3}{2}e^{2x} – \frac{1}{2} – x\)

The exponential expression \(e^{2x}\) is defined for every real number \(x\).

The terms \(-\frac{1}{2}\) and \(-x\) are also defined for every real number \(x\).

So there are no restrictions on \(x\).

Therefore, the domain is:

\((-\infty, \infty)\)

Step 4: Find the y-intercept

The y-intercept occurs when \(x = 0\).

Substitute \(x = 0\) into the function:

\(y(0) = \frac{3}{2}e^{2(0)} – \frac{1}{2} – 0\)

Simplify the exponent:

\(2(0) = 0\)

So:

\(y(0) = \frac{3}{2}e^0 – \frac{1}{2}\)

Since:

\(e^0 = 1\)

we get:

\(y(0) = \frac{3}{2}(1) – \frac{1}{2}\)

Simplify:

\(y(0) = \frac{3}{2} – \frac{1}{2}\)

\(y(0) = \frac{2}{2}\)

\(y(0) = 1\)

So the y-intercept is:

\((0, 1)\)

Step 5: Find the first derivative

The first derivative tells us where the function increases or decreases.

Start with:

\(y(x) = \frac{3}{2}e^{2x} – \frac{1}{2} – x\)

Differentiate each term separately.

First term:

\(\frac{3}{2}e^{2x}\)

To differentiate \(e^{2x}\), use the chain rule.

The derivative of \(e^{2x}\) is:

\(2e^{2x}\)

So:

\(\frac{d}{dx}\left(\frac{3}{2}e^{2x}\right) = \frac{3}{2} \cdot 2e^{2x}\)

Simplify:

\(\frac{3}{2} \cdot 2 = 3\)

So the derivative of the first term is:

\(3e^{2x}\)

Second term:

\(-\frac{1}{2}\)

This is a constant, so its derivative is:

0

Third term:

\(-x\)

The derivative of \(-x\) is:

\(-1\)

Now combine the derivatives:

\(y'(x) = 3e^{2x} – 1\)

Step 6: Find critical points

Critical points occur where the first derivative equals zero or where the first derivative is undefined.

The derivative is:

\(y'(x) = 3e^{2x} – 1\)

This derivative is defined for every real number \(x\), so we only need to solve:

\(3e^{2x} – 1 = 0\)

Add \(1\) to both sides:

\(3e^{2x} = 1\)

Divide both sides by \(3\):

\(e^{2x} = \frac{1}{3}\)

Take the natural logarithm of both sides:

\(\ln(e^{2x}) = \ln\left(\frac{1}{3}\right)\)

Use the identity \(\ln(e^a) = a\):

\(2x = \ln\left(\frac{1}{3}\right)\)

Since:

\(\ln\left(\frac{1}{3}\right) = -\ln 3\)

we get:

\(2x = -\ln 3\)

Divide both sides by \(2\):

\(x = -\frac{\ln 3}{2}\)

So the function has one critical point at:

\(x = -\frac{\ln 3}{2}\)

Step 7: Find the y-value at the critical point

Now substitute \(x = -\frac{\ln 3}{2}\) into the original function.

The function is:

\(y(x) = \frac{3}{2}e^{2x} – \frac{1}{2} – x\)

Substitute:

\(y\left(-\frac{\ln 3}{2}\right) = \frac{3}{2}e^{2\left(-\frac{\ln 3}{2}\right)} – \frac{1}{2} – \left(-\frac{\ln 3}{2}\right)\)

Simplify the exponent:

\(2\left(-\frac{\ln 3}{2}\right) = -\ln 3\)

So:

\(y\left(-\frac{\ln 3}{2}\right) = \frac{3}{2}e^{-\ln 3} – \frac{1}{2} + \frac{\ln 3}{2}\)

Now simplify \(e^{-\ln 3}\).

Since:

\(e^{\ln 3} = 3\)

we have:

\(e^{-\ln 3} = \frac{1}{3}\)

So:

\(y\left(-\frac{\ln 3}{2}\right) = \frac{3}{2}\cdot \frac{1}{3} – \frac{1}{2} + \frac{\ln 3}{2}\)

Simplify the first product:

\(\frac{3}{2}\cdot \frac{1}{3} = \frac{1}{2}\)

So:

\(y\left(-\frac{\ln 3}{2}\right) = \frac{1}{2} – \frac{1}{2} + \frac{\ln 3}{2}\)

\(y\left(-\frac{\ln 3}{2}\right) = \frac{\ln 3}{2}\)

So the critical point is:

\(\left(-\frac{\ln 3}{2}, \frac{\ln 3}{2}\right)\)

Step 8: Determine whether the critical point is a minimum or maximum

To classify the critical point, find the second derivative.

The first derivative is:

\(y'(x) = 3e^{2x} – 1\)

Differentiate again:

\(y”(x) = 6e^{2x}\)

Since \(e^{2x}\) is always positive for every real number \(x\), we know:

\(6e^{2x} > 0\)

So:

\(y”(x) > 0\)

That means the function is concave up everywhere.

Therefore, the critical point is a minimum.

The minimum point is:

\(\left(-\frac{\ln 3}{2}, \frac{\ln 3}{2}\right)\)

Step 9: Find where the function increases and decreases

The first derivative is:

\(y'(x) = 3e^{2x} – 1\)

The critical value is:

\(x = -\frac{\ln 3}{2}\)

When \(x < -\frac{\ln 3}{2}\), the value of \(e^{2x}\) is less than \(\frac{1}{3}\), so:

\(3e^{2x} – 1 < 0\)

Therefore, the function is decreasing on:

\(\left(-\infty, -\frac{\ln 3}{2}\right)\)

When \(x > -\frac{\ln 3}{2}\), the value of \(e^{2x}\) is greater than \(\frac{1}{3}\), so:

\(3e^{2x} – 1 > 0\)

Therefore, the function is increasing on:

\(\left(-\frac{\ln 3}{2}, \infty\right)\)

Step 10: Find the range

The function has a minimum value at:

\(x = -\frac{\ln 3}{2}\)

The minimum value is:

\(y = \frac{\ln 3}{2}\)

Because the function is concave up everywhere and grows without bound as \(x\) becomes large, the range is:

\(\left[\frac{\ln 3}{2}, \infty\right)\)

Final answer

The function is:

\(y(x) = \frac{3}{2}e^{2x} – \frac{1}{2} – x\)

Its domain is:

\((-\infty, \infty)\)

Its y-intercept is:

\((0, 1)\)

Its first derivative is:

\(y'(x) = 3e^{2x} – 1\)

Its only critical point is:

\(\left(-\frac{\ln 3}{2}, \frac{\ln 3}{2}\right)\)

This point is a minimum.

The function decreases on:

\(\left(-\infty, -\frac{\ln 3}{2}\right)\)

The function increases on:

\(\left(-\frac{\ln 3}{2}, \infty\right)\)

The range is:

\(\left[\frac{\ln 3}{2}, \infty\right)\)