To determine the intervals where the function $f(x) = \frac{e^{4x^2 - 2x}}{4x}$ is increasing, we need to follow these steps:

Step 1: Compute the derivative $f'(x)$.

We use the quotient rule: $\frac{d}{dx} \left( \frac{u(x)}{v(x)} \right) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2},$ where: $u(x) = e^{4x^2 - 2x}, \quad v(x) = 4x.$

• Derivative of the numerator ($u(x)$): $u'(x) = e^{4x^2 - 2x} \cdot \frac{d}{dx}(4x^2 - 2x) = e^{4x^2 - 2x} \cdot (8x - 2).$

• Derivative of the denominator ($v(x)$): $v'(x) = \frac{d}{dx}(4x) = 4.$

Using the quotient rule, the derivative is: [ f'(x) = \frac{\left[ e^{4x^2 - 2x} \cdot (8x - 2) \cdot (4x) \right] - \left[ e^{4x^2 - 2x} \cdot 4 \right]}{(4x)^2}. ]

Step 2: Simplify the expression for $f'(x)$.

Factoring out $e^{4x^2 - 2x}$ from the numerator: $f'(x) = \frac{e^{4x^2 - 2x} \left[ (8x - 2)(4x) - 4 \right]}{16x^2}.$

Expand and simplify the numerator: $(8x - 2)(4x) = 32x^2 - 8x,$ so: $(32x^2 - 8x) - 4 = 32x^2 - 8x - 4.$

Thus: $f'(x) = \frac{e^{4x^2 - 2x} (32x^2 - 8x - 4)}{16x^2}.$

Step 3: Analyze where $f'(x) > 0$.

Since $e^{4x^2 - 2x} > 0$ for all $x$, the sign of $f'(x)$ depends on: $32x^2 - 8x - 4 > 0.$

Factorize or solve the quadratic equation $32x^2 - 8x - 4 = 0$ using the quadratic formula: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}, \quad a = 32, \, b = -8, \, c = -4.$

$x = \frac{-(-8) \pm \sqrt{(-8)^2 - 4(32)(-4)}}{2(32)} = \frac{8 \pm \sqrt{64 + 512}}{64} = \frac{8 \pm \sqrt{576}}{64}.$

$x = \frac{8 \pm 24}{64}.$

This gives two roots: $x = \frac{32}{64} = \frac{1}{2}, \quad x = \frac{-16}{64} = -\frac{1}{4}.$

Step 4: Determine intervals of increase.

The critical points are $x = -\frac{1}{4}$ and $x = \frac{1}{2}$. Test the sign of $32x^2 - 8x - 4$ in each interval:

1. $x < -\frac{1}{4}$,
2. $-\frac{1}{4} < x < \frac{1}{2}$,
3. $x > \frac{1}{2}$.

• For $x < -\frac{1}{4}$, choose $x = -1$: $32(-1)^2 - 8(-1) - 4 = 32 + 8 - 4 = 36 > 0.$

• For $-\frac{1}{4} < x < \frac{1}{2}$, choose $x = 0$: $32(0)^2 - 8(0) - 4 = -4 < 0.$

• For $x > \frac{1}{2}$, choose $x = 1$: $32(1)^2 - 8(1) - 4 = 32 - 8 - 4 = 20 > 0.$

Thus, $f'(x) > 0$ on: $(-\infty, -\frac{1}{4}) \quad \text{and} \quad (\frac{1}{2}, \infty).$

Final Answer:

The function $f(x)$ is increasing on the intervals: $(-\infty, -\frac{1}{4}) \cup (\frac{1}{2}, \infty).$

Would you like a deeper explanation of any step or a graph of the function? 😊

5 Related Questions:

1. How do you determine where a function is decreasing?
2. What is the significance of critical points in calculus?
3. How is the quadratic formula applied to solve inequalities?
4. Why is the exponential function always positive?
5. What role does the derivative test play in analyzing functions?

Tip:

Always simplify the derivative as much as possible before analyzing its sign—it makes finding intervals of increase or decrease much easier!