Question:

Let $f(x) = 3x^2 - 2x + 1$.

(a) Compute the derivative $f'(x)$ using the definition of the derivative.

(b) Find the equation of the tangent line to the graph of $f(x)$ at the point $(2, f(2))$.

(c) Find the value of $f'(2)$.

Answer:

(a) To compute the derivative $f'(x)$ using the definition of the derivative, we recall that:

Let's substitute $f(x) = 3x^2 - 2x + 1$ into this formula and simplify:

Expanding and collecting like terms:

Cancelling out terms:

Simplifying:

Taking the limit as $h$ approaches 0:

Therefore, the derivative $f'(x)$ of $f(x) = 3x^2 - 2x + 1$ is $6x - 2$.

(b) To find the equation of the tangent line to the graph of $f(x)$ at the point $(2, f(2))$, we need both the slope of the tangent line and a point on the line.

Since $f'(x) = 6x - 2$, the slope of the tangent line at $x = 2$ is $f'(2) = 6(2) - 2 = 10$.

To find a point on the tangent line, we substitute $x = 2$ into $f(x)$:

Therefore, the point on the tangent line is $(2, 11)$.

Using the point-slope form of a linear equation, we have:

Substituting $m = 10$, $x_1 = 2$, and $y_1 = 11$:

Simplifying:

Hence, the equation of the tangent line to the graph of $f(x)$ at the point $(2, f(2))$ is $y = 10x - 9$.

(c) To find the value of $f'(2)$, we substitute $x = 2$ into the derivative $f'(x) = 6x - 2$:

Therefore, the value of $f'(2)$ is 10.