Question

let (f(x)=12x + 3-3e^{x}). then the equation of the tangent line to the graph of (f(x)) at the point ((0,0)) is given by (y = mx + b) for (m=square) (b=square)

Explanation:

Step1: Find the derivative of f(x)

The derivative of $f(x)=12x + 3-3e^{x}$ using the sum - rule and derivative formulas. The derivative of $12x$ is $12$, the derivative of a constant $3$ is $0$, and the derivative of $-3e^{x}$ is $-3e^{x}$. So $f^\prime(x)=12 - 3e^{x}$.

Step2: Calculate the slope m

The slope $m$ of the tangent line at the point $(x_0,y_0)=(0,0)$ is $f^\prime(x_0)$. Substitute $x = 0$ into $f^\prime(x)$: $m=f^\prime(0)=12-3e^{0}=12 - 3\times1=9$.

Step3: Find the y - intercept b

The equation of the line is $y=mx + b$. We know that the line passes through the point $(0,0)$ and $m = 9$. Substitute $x = 0$, $y = 0$ and $m = 9$ into $y=mx + b$: $0=9\times0 + b$, so $b = 0$.

Answer:

$m = 9$
$b = 0$