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FrozenT [24]
3 years ago
9

Use a table of function values to approximate an x-value in which the exponential function exceeds the polynomial function. f(x)

= 5x + 4 h(x) = x2 + 8x + 24
x = -3
x = 3
x = 4
x = 5
Mathematics
2 answers:
WINSTONCH [101]3 years ago
6 0
The exponential function is f(x)= 5^{x} +4
The polynomial function is h(x)= x^{2} +8x+24 

We want a value of x as such that f(x)\ \textgreater \ h(x), so

5^{x}+4 > x^{2} +8x+24

We are provided with four possible value of x. By process of trial and error, we will narrow down two values of x when the statement is changing from false to true

first, we try x=-3 by substituting into f(x) and h(x)

f(-3)=5^{-3}+4 = 4.008 and h(-3)= (-3)^{2}+8(-3)+24=9
so f(x)>h(x) which is not the inequality wanted

We try x=3 by substituting into f(x) and h(x)
f(3)= 5^{3}+4=129 and h(3)= (3)^{2}+8(3)+24=57, 
so f(x) > h(x) as wanted

We know that any values of x greater than three will give f(x) greater than h(x), so we narrow down to two values that make the statement changes from false to true.

There is one value between -3 and three that makes f(x) greater than h(x). From here we use the method of trial and error. 

We can start with the mid value between -3 and three which is 0
f(0)= 5^{0}+4=5
h(0)= 0^{2}+8(0)+24=24
f(x) is less than h(x) which isn't the inequality we want so here we narrow down the value of x must be between 0 and 3

We can try mid-value between 0 and three which is 1.5
f(1.5)= 5^{1.5}+4=15.18
h(1.5)= (1.5)^{2}+8(1.5)+24=38.25
f(x) is still less than h(x), so we now narrow down the value of x between 1.5 and 3

We try again using the mid-value between 1.5 and three which is 2.25
f(2.25)= 5^{2.25}+4=41.38
h(2.25)= 2.25^{2}+8(2.25)+24=47.0625
We still have f(x) less than h(x) so we can narrow down further the value of x between 2.25 and 3 (which is quite a short interval already)

We can keep trying by choosing a value of x says x=2.6
f(2.6)= 5^{2.6}+4=69.66
h(2.6)= 2.6^{2}+8(2.6)+24=51.56
We have f(x) greater then h(x) 

The narrowing down process continues where we now have the interval of x between 2.25 and 2.6. Keeping to one decimal place we can find the 
value of x=2.4 is the approximate value where f(x) is greater than h(x).

Answer: x=2.4
enyata [817]3 years ago
5 0

The correct answer is x = 5.

Trust me, I just did the quiz.

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To solve this problem, we use the formula:

z = (x – u) / s

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4 years ago
The area if a rectangle is 53.6cm^2.If the length is multiplied by four and the width is halved, the area would then be?
e-lub [12.9K]

Answer:

26.8cm²

Step-by-step explanation:

the area if a rectangle is 53.6cm^2.If the length is multiplied by four and the width is halved, the area would then be?​

Area of rectangle = length x width

53.6 = 4L x 1/2 W

53.6 = 2L × W

Divide both sides by 2

L × W = 53.6 ÷ 2

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Answer:

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