Question

Consider the function f(x) = x3 − 4x2 + 2. Calculate the limit of the difference quotient at x0 = 3 for f(x).

Answer 1

Part 1.

Answer: $\boxed{f'(x)=3x^2-8x}$

The limit of the difference quotient is simply the derivative, so we can express this as follows:

$f'(x)=\underset{\Delta x\rightarrow0}{lim}\frac{\triangle y}{\Delta x}=\frac{f(x+\Delta x)-f(x)}{\Delta x}$

So our function is:

$f(x)=x^3-4x^2+2$

Taking the derivative, we have:

$f'(x)=3x^2-8x$

So the correct option is:

$\boxed{f'(x)=3x^2-8x}$

Part 2.

Answer: $\boxed{y=3x-16}$

The equation of the line that passes through the same point can be found as:

$y-y_{0}=m(x-x_{0})$

Where $x_{0}=3$, so we need to find $y_{0}$. Plug in that x-value in the function we have:

$y_{0}=f(3)=(3)^3-4(3)^2+2 \\ \\ y_{0}=-7 \\ \\ \\ So \ the \ point \ is: \\ \\ P(x_{0},y_{0})=(3,-7)$

And the slope is:

$m=f'(3)=3(3)^2-8(3) \\ \\ m=3$

Then, the equation of the line is:

$y-(-7)=3(x-3) \\ \\ \therefore y+7=3x-9 \\ \\ \boxed{y=3x-16}$

Part 3.

Answer: Shown below

As you can see below, the graph of the function of $f$ is continuous. This is so because we have plotted a polynomial function whose domain is the set of all real numbers. So the function is defined at the point $P(3,-7)$, so the derivative exists at this point, hence we can calculate a tangent line there. In conclusion, we get the graph shown below. The blue line is the tangent line while the red curve is the graph of $f$