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rosijanka [135]
3 years ago
7

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

Mathematics
1 answer:
m_a_m_a [10]3 years ago
3 0
<h2>Part 1.</h2>

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}

<h2>Part 2.</h2>

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}

<h2 /><h2>Part 3.</h2>

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

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