Fis a twice differentiable function that is defined for all reals. the value of f "(x) is given for several values of x in the table below. x –8 –3 0 3 8 f " (x) –4 –2 0 4 5 if f " (x) is always increasing, which statement about f(x) must be true? f(x) is concave downwards for all x. f(x) passes through the origin. f(x) has a relative minimum at x = 0. < - my answer f(x) has a point of inflection at x = 0.

The correct answer is actually the last one.

The second derivative gives us information about the concavity of a function: if then the function is concave downwards in that point, whereas if then the function is concave upwards in that point.

This already shows why the first option is wrong - if the function was concave downwards for all x, then the second derivate would have been negative for all x, which isn't the case, because we have, for example,

Also, the second derivative gives no information about specific points of the function. Suppose, in fact, that passes through the origin, so . Now translate the function upwards, for example. we have that doesn't pass through the origin, but the second derivative is always . So, the second option is wrong as well.

Now, about the last two. The answer you chose would be correct if the exercise was about the first derivative . In fact, the first derivative gives information about the increasing or decreasing behaviour of the function - positive and negative derivative, respectively. So, if the first derivative is negative before a certain point and positive after that point. It means that the function is decreasing before that point, and increasing after. So, that point is a relative minimum.

But in this exercise we're dealing with second derivative, so we don't have information about the increasing/decreasing behaviour. Instead, we know that the second derivative is negative before zero - which means that the function is concave downwards before zero - and positive after zero - which means that the function is concave upwards after zero.

A point where the function changes its concavity is called a point of inflection, which is the correct answer.

∠1≅∠2 by the alternate exterior angles theorem.

step-by-step explanation:

given, a ∥ b and ∠1 ≅ ∠3 .we have to prove that e ∥ f

we know that ∠1≅∠3 and that a || b because they are given. we see that by the alternate exterior angles theorem. therefore, ∠2≅∠3 by the transitive property. so, we can conclude that e || f by the converse alternate exterior angles theorem.

we have to fill the missing statement.

transitivity property states that if a = b and b = c, then a = c.

now, given ∠1≅∠3 and by transitivity property ∠2≅∠3 .

hence, to apply transitivity property one angle must be common which is not in result after applying this property which is ∠1.

the only options in which ∠1 is present are ∠1 and ∠2, ∠1 and ∠4

∠1 and ∠4 is not possible ∵ after applying transitivity we didn't get ∠4.

hence, the missing statement is ∠1≅∠2.

so, ∠1≅∠2 by the alternate exterior angles theorem.