Find the approximate value of f(4.04), where f(x)=7x³+6x²-4x+3

The correct option is 546.2.

To find the approximate value of $f\left(4.04\right)$ for the function $f\left(x\right)=7x^3+6x^2-4x+3$, we can use the concept of differentials or linear approximation.

The linear approximation of a function $f\left(x\right)$ near a point $x=a$ is given by the formula:
$f(a+\Delta x)\approx f\left(a\right)+f^{\prime }\left(a\right)\Delta x$

Here, we choose:
$a=4$
$\Delta x=0.04$
So, $x=a+\Delta x=4+0.04=4.04$.

First, let's calculate $f\left(a\right)=f\left(4\right)$:
$f\left(4\right)=7\left(4^3\right)+6\left(4^2\right)-4\left(4\right)+3$
$f\left(4\right)=7\left(64\right)+6\left(16\right)-16+3$
$f\left(4\right)=448+96-16+3$
$f\left(4\right)=531$

Next, we find the derivative of $f\left(x\right)$:
$f^{\prime }\left(x\right)=\frac{d}{dx}(7x^3+6x^2-4x+3)$
$f^{\prime }\left(x\right)=21x^2+12x-4$

Now, calculate $f^{\prime }\left(a\right)=f^{\prime }\left(4\right)$:
$f^{\prime }\left(4\right)=21\left(4^2\right)+12\left(4\right)-4$
$f^{\prime }\left(4\right)=21\left(16\right)+48-4$
$f^{\prime }\left(4\right)=336+48-4$
$f^{\prime }\left(4\right)=380$

Substitute these values into the linear approximation formula:
$f\left(4.04\right)\approx f\left(4\right)+f^{\prime }\left(4\right)\Delta x$
$f\left(4.04\right)\approx 531+380\times 0.04$
$f\left(4.04\right)\approx 531+15.2$
$f\left(4.04\right)\approx 546.2$

Thus, the approximate value of $f\left(4.04\right)$ is 546.2.