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2 months ago

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

Mathematics

1 answer:

tester [12.3K]2 months ago

3 0

Part 1.

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

The limit of the difference quotient equates to the derivative, which can be expressed as follows:

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

Thus, our function is:

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

Calculating the derivative results in:

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

Consequently, the appropriate option is:

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

Part 2.

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

The equation of the line that goes through the same point can be determined as:

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

Where $x_{0}=3$ , requiring us to find $y_{0}$ . Substitute that x-value into the function we have:

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

The slope calculated is:

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

Therefore, the equation of the line is:

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

Part 3.

Answer: As illustrated below

As demonstrated below, the function's graph for $f$ is continuous. This occurs because we plotted a polynomial function whose domain encompasses all real numbers. Consequently, the function is defined at the point $P(3,-7)$ , ensuring the derivative is valid at this location, thereby allowing us to calculate the tangent line there. In summary, we present the graph below. The blue line indicates the tangent line while the red curve represents the graph of $f$ .

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