Find the intervals in which f(x) = 2x² – 3x is increasing

The correct option is (3/4, ∞).

To find the intervals in which the function $f\left(x\right)=2x^2-3x$ is increasing, we need to determine where its first derivative is strictly positive (i.e., $f^{\prime }\left(x\right)>0$).

Step 1: Find the derivative of the function.
Using the power rule of differentiation, we find $f^{\prime }\left(x\right)$:
$$f^{\prime }\left(x\right)=\frac{d}{dx}(2x^2-3x)=4x-3$$

Step 2: Set up the inequality for an increasing function.
A function is increasing on an interval where its derivative is positive:
$$f^{\prime }\left(x\right)>0$$
Substituting the derivative we found:
$$4x-3>0$$

Step 3: Solve the inequality for x.
Add 3 to both sides of the inequality:
$$4x>3$$
Divide both sides by 4:
$$x>\frac{3}{4}$$

Thus, the function is increasing for all values of $x$ greater than $\frac{3}{4}$.
In interval notation, this is represented as (3/4, ∞).