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bazaltina [42]
3 years ago
5

Use the Intermediate Value Theorem to show that the polynomial f(x)=x^3+x^2-2x+5 has a real zero between -3 and -1.

Mathematics
1 answer:
Triss [41]3 years ago
6 0

You need to note first two things.

1. The function f(x) need to be a continuous function on [-3,-1] (indeed f(x) is a polynomial function, is continuous on every real number x)

2. Take the closed interval [-3,-1]

Now, evaluate f(-3) and f(-1).

f(-3) = (-3)^3 + (-3)^2 -2(-3) + 5 = -27 + 9 +6 + 5 = -7 < 0\\f(-1) = (-1)^3 + (-1)^2 -2 (-1) + 5 = -1 + 1 +2 + 5 = 7 >0

So, we have f(-3) < 0  and f(-1). So,  f(-3) < 0 < f(-1), then by the Intermediate Value Theorem, there exist c \in (-3,-1) such that f(c) = 0.

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The average rate of change of function g(x)=5(2)^{x} from x = 3 to x = 4 is 4 times that from x = 1 to x = 2.

The correct option is (A).

What is the average rate of change of a function?

The average rate at which one quantity changes in relation to another's change is referred to as the average rate of change function.

Using function notation, we can define the Average Rate of Change of a function f from a to b as:

                                     rate(m) = \frac{f(b)-f(a)}{b-a}

The given function is  g(x) = 5(2)^{x},

Now calculating the average rate of change of function from x = 1 to x = 2.

                               m = \frac{g(b)-g(a)}{b-a}\\ m = \frac{g(2)-g(1)}{2-1}\\\\m=\frac{5(2)^{2}-5(2)^1}{2-1}\\ m=\frac{10}{1} \\m=10

Now, calculate the average rate of change of function from x = 3 to x = 4.

                                 m = \frac{g(b)-g(a)}{b-a}\\ m = \frac{g(4)-g(3)}{4-3}\\\\m=\frac{5(2)^{4}-5(2)^3}{4-3}\\ m=\frac{40}{1} \\m=40

The jump from m = 10 to m = 40 is "times 4".

So option (A) is correct.

Hence, The average rate of change of function g(x)=5(2)^{x} from x = 3 to x = 4 is 4 times that from x = 1 to x = 2.

To learn more about the average rate of change of function, visit:

brainly.com/question/24313700

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