The correct option is Option D: f(x)=(5)ˣ.
The function is increasing where its derivative is greater than 0.
for example, f(x) is increasing then f'(x) is greater than 0.
Here f(x) = aˣ
so the dervative of the function
d(f(x))/dx=d(aˣ)/dx
⇒f'(x)= ㏑(a)aˣ
f'(x)>0
⇒㏑(a)aˣ >0
as aˣ is always greater than 0,
⇒㏑(a)>0
⇒a>1
as ㏑(a) is greater than 0 when a >1 and is lesser than when 0<a<1.
checking all the options
Option A. f(x)=(1/5)ˣ here a =1/5<1 so, f(x) is not increasing. Option A is incorrect.
Option B: f(x)=(0.5)ˣ here a =0.5<1 so, f(x) is not increasing. Option B is incorrect.
Option C: f(x)=(1/15)ˣ here a =1/15<1 so, f(x) is not increasing. Option C is incorrect.
Option D: f(x)=(5)ˣ here a =5>1 so, f(x) is increasing. Option A is correct.
Therefore the correct option is Option D: f(x)=(5)ˣ.
Learn more about increasing function
here: brainly.com/question/1503051
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The 3 consecutive integers of a sum of 81<span>X+(X+2)+(X+4)=81
3X+6=81
3X=81-6
3X=75
X=75/3</span>
Answer:
C. >
F. ≠
Step-by-step explanation:
C. 23 is greater than 16
F. 23 is not equal to 16
Answer:
<em>The temperature of both appliances is 70° after 6 hours.</em>
<em>At 6 hours, both temperatures will be equal</em>
Step-by-step explanation:
<u>Equations</u>
Let's call
T1 = Temperature inside of the refrigerator
T2 = Temperature inside of the oven.
x = Time in hours elapsed since the power goes out.
We know the temperature of the refrigerator rises at 5° per hour. Thus the function that models its temperature is:
T1 = 40 + 5x
We are also given the oven temperature drops by 45° per hour, thus:
T2 = 340 - 45x
a) Find the value of x for both temperatures to be the same.
We equal T1 and T2:
40 + 5x = 340 - 45x
Adding 45x
40 + 5x + 45x = 340
Subtracting 40:
5x + 45x = 340 - 40
Operating:
50x = 300
x = 300/50
x = 6
At 6 hours, both temperatures will be equal
b)
The common temperature can be found by substituting x in any function:
T1 = 40 + 5x = 40 + 5(6) = 40 + 30 = 70
T2 = 340 - 45x = 340 - 45(6) = 70
The temperature of both appliances is 70° after 6 hours.