Find the approximate value of f(5.03), where f(x)=4x²-7x+2

The correct option is 67.99.

To find the approximate value of $f\left(5.03\right)$ for the function $f\left(x\right)=4x^2-7x+2$, we can use the concept of differentials and linear approximation.

Let $x=5$ and the change in $x$ be $∆x=dx=0.03$.

The differential of $y=f\left(x\right)$ is given by:
$dy=f^{\prime }\left(x\right)dx$

First, let's find the derivative of $f\left(x\right)$:
$f^{\prime }\left(x\right)=\frac{d}{dx}(4x^2-7x+2)=8x-7$

Now, evaluate $f\left(x\right)$ and $f^{\prime }\left(x\right)$ at $x=5$:
$f\left(5\right)=4\left(5^2\right)-7\left(5\right)+2=4\left(25\right)-35+2=100-35+2=67$
$f^{\prime }\left(5\right)=8\left(5\right)-7=40-7=33$

Using the differential formula to approximate the change in the function value:
$dy=f^{\prime }\left(5\right)dx=33\times 0.03=0.99$

Therefore, the approximate value of $f\left(5.03\right)$ is:
$f\left(5.03\right)\approx f\left(5\right)+dy=67+0.99=67.99$