(a) The graph of f has a point of inflection somewhere between x = -1 and x= 3

(b) f’(-1) = 0

(c) this is wrong

(d) The graph of f has a horizontal tangent line at x = 3

(e) The graph of f intersects both axes - Correct answer

I understand why e is correct, but I do not get why a, b, and d are all wrong. Aren’t b and d to be expected since they are relative max/min’s? I also can’t imagine a case in which "a" is incorrect. Can someone explain why they are wrong? Thank you!

\( f’(x)=k(x+1)(x-3) \)

\( f’(x)=k(x^2-2x-3) \)

\( f(x)=k( \frac{x^3}{3}-x^2-3x )+C \)

\( f"(x)=k(2x-2) \)

if f"(x)=0 then x=1

therefore a) is true

f’(-1)=0 b) is also true

f’(3)=0 d) is also true

e) option is also correct

i think

First of all we need to review the concept of concavity. So, this is related to the second derivative. If we want to think about f double prime, then we need to think about how f prime changes, how the slopes of the tangent lines change.

So:

**1) **On intervals where \( f’’>0 \), the function is **concave up **(Depicted in bold purple in Figure 1)

**2) **On intervals where \( f’’<0 \), the function is **concave down **(Depicted in bold green in Figure 1)

Points where the graph of a function changes from concave up to concave down, or vice versa, are called **inflection points. **

Suppose we have a function whose graph is shown below. Therefore we have:

**(a) The graph of f has a point of inflection somewhere between x = -1 and x= 3**

This is **true. **As you can see from the Figure 2 the **inflection point **is pointed out in green. In this point the function changes** from concave down to concave up. **

**(b) f’(-1) = 0 **

This is **true. **In \( x=-1 \) there’s a **maximum point. ** In this point the slope of the tangent line is in fact zero, that is, the function has an horizontal line as shown in Figure 3 (the line in green).

**(c) this is wrong **

This is **false **because we have demonstrated that the previous statement are true.

**(d) The graph of f has a horizontal tangent line at x = 3**

This is **true. **As in case (b) the function has an horizontal line as shown in Figure 3 (the line in orange) because in \( x=3 \) there is a **minimum point. **

**(e) The graph of f intersects both axes **

This is **true **according to **Bolzano’s Theorem. **Apaticular case of the the **Intermediate Value Theorem** is the Bolzano’s theorem. Suppose that \( f(x) \) is a continuous function on a closed interval \( [a, b] \) and takes the values of the opposite sign at the extremes, and there is at least one \( c \in (a, b) \ such \ that \ f(c)=0 \)