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Ne4ueva [31]
3 years ago
11

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 t

he 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!

Mathematics
2 answers:
miv72 [106K]3 years ago
8 0

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

evablogger [386]3 years ago
3 0
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
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