Question

determine the equation of the line l that is tangent to the polynomial f at the point (4, f(4)). where f(x)=5x^3 - 42x^2 + 98x - 37

Explanation:

Step1: Find the derivative of $f(x)$

Using the power - rule $\frac{d}{dx}(x^n)=nx^{n - 1}$, we have $f'(x)=15x^{2}-84x + 98$.

Step2: Evaluate $f(4)$

Substitute $x = 4$ into $f(x)$:
$f(4)=5\times4^{3}-42\times4^{2}+98\times4 - 37=5\times64-42\times16 + 392-37=320-672+392-37=3$.

Step3: Evaluate $f'(4)$

Substitute $x = 4$ into $f'(x)$:
$f'(4)=15\times4^{2}-84\times4 + 98=15\times16-336 + 98=240-336 + 98=2$.

Step4: Use the point - slope form $y - y_1=m(x - x_1)$

Here, $(x_1,y_1)=(4,3)$ and $m = f'(4)=2$.
The equation of the tangent line is $y - 3=2(x - 4)$.
Expand to get $y=2x-8 + 3=2x-5$.

Answer:

$y = 2x-5$