What will be the increment of the differentiable function f(x) = 2x2 – 3x + 2 when x changes from 3.02 to 3?

The correct option is 0.18.

To find the increment (or differential) of the differentiable function $f\left(x\right)=2x^2-3x+2$, we can use the concept of differentials. The differential $dy$ (which approximates the actual change in the function value, $\Delta y$) is given by the formula:
$dy=f^{\prime }\left(x\right)dx$
where $f^{��}\left(x\right)$ is the first derivative of the function and $dx$ represents the change in the independent variable $x$.

Step 1: Determine the change in $x$ ($dx$)
Here, the value of $x$ changes between $3$ and $3.02$. The magnitude of the change in $x$ is:
$dx=3.02-3=0.02$

Step 2: Find the derivative of the function $f\left(x\right)$
We differentiate $f\left(x\right)=2x^2-3x+2$ with respect to $x$ using the power rule:
$f^{\prime }\left(x\right)=\frac{d}{dx}(2x^2-3x+2)=4x-3$

Step 3: Evaluate the derivative at the base point $x=3$
Substituting $x=3$ into the derivative expression:
$f^{\prime }\left(3\right)=4\left(3\right)-3=12-3=9$

Step 4: Calculate the increment $dy$
Multiply the derivative value by $dx$ to find the approximate increment of the function:
$dy=f^{\prime }\left(3\right)·dx=9·0.02=0.18$

Therefore, the increment of the function is 0.18.