Question

If a function f is continuous for all x and if f has a relative maximum at (-1, 4) and a relative minimum at (3, -2), which of the following statements must be true? 
(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!

Answer 1

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 $f''$, 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 $c \in (a,b) \ such \ that \ f(c)=0$

Answer 2

$f'(x)=k(x+1)(x-3)$
$f'(x)=k(x^2-2x-3)$
$f(x)=k( \frac{x^3}{3}-x^2-3x )+C$
$f"(x)=k(2x-2)$
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