If f(1) = 13 and f '(x) ≥ 3 for 1 ≤ x ≤ 3, how small can f(3) possibly be?

The answer to this question is 19.

Step-by-step explanation:

Given that :

f(x)=13.

f'(x)=3.   1 ≤ x ≤ 3.

Integrate

∫f'(x) dx=∫3 dx

f(x)=3x+c        1 ≤ x ≤ 3.

f(1)= 3+c  

c=13-3 =10.

f(x)=3x+10  1 ≤ x ≤ 3.

now ,

f(3)=3(3)+10=19.

So f(3) is at least 19.

last option

step-by-step explanation:

given that δdef is right and isoceles,

assuming that side df is the hypotenuse means that de an ef are the other 2 sides that intersect at 90° and are of equal length.

because they are of equal length, we can say that they are congruent

because they intersect at 90°, their slopes must be opposite reciprocals.

only the last option satisfies these conditions.