Question

1. a companys profit function is: $p(x) = -3x^2 + 60x - 200$ find the marginal profit at $x = 6$. interpret your answer.

1. given the function:

$$f(x) = \
LATEXBLOCK0
$$

find:

• $\lim\limits_{x \to 1^-} f(x)$
• $\lim\limits_{x \to 1^+} f(x)$
• $\lim\limits_{x \to 1} f(x)$

Explanation:

Question 3

Step1: Recall Marginal Profit Definition

Marginal profit is the derivative of the profit function. So, first, find \( P'(x) \) for \( P(x) = -3x^2 + 60x - 200 \).
Using the power rule \( \frac{d}{dx}(ax^n)=nax^{n - 1} \), we have:
\( P'(x)=\frac{d}{dx}(-3x^2)+\frac{d}{dx}(60x)-\frac{d}{dx}(200) \)
\( P'(x)=-6x + 60-0=-6x + 60 \)

Step2: Evaluate at \( x = 6 \)

Substitute \( x = 6 \) into \( P'(x) \):
\( P'(6)=-6(6)+60 \)
\( P'(6)=-36 + 60 \)
\( P'(6)=24 \)

Interpretation:

The marginal profit at \( x = 6 \) is 24. This means that when the company produces and sells the 6th unit (or at the production level of \( x = 6 \)), the profit is increasing at a rate of 24 units of currency (e.g., dollars) per unit increase in production (or sales) at that point.

When \( x
ightarrow1^{-} \), \( x < 1 \), so we use the piece \( f(x)=2x + 3 \).
Substitute \( x = 1 \) into \( 2x + 3 \):
\( \lim_{x
ightarrow1^{-}}f(x)=2(1)+3=2 + 3 = 5 \)

When \( x
ightarrow1^{+} \), \( x\geq1 \), so we use the piece \( f(x)=x^2 + 1 \).
Substitute \( x = 1 \) into \( x^2 + 1 \):
\( \lim_{x
ightarrow1^{+}}f(x)=1^2 + 1=1 + 1 = 2 \)

Answer:

(Marginal Profit Value):
\( 24 \)

Question 4

Part 1: \( \lim_{x

ightarrow1^{-}}f(x) \)