Verify that the intermediate value theorem applies to the indicated interval and find the value of c guaranteed by the theorem. f(x) = x2 7x 1, [0, 9], f(c) = 19 c =

Sagot :

we verified the intermidiate value theorem applies to the function f(x) = x^2 + 7x  + 1 . And the value of c is 2.

According to the given question.

We have a function.

f(x) = x^2 + 7x  + 1

As, we know that "the Intermediate Value Theorem (IVT) states that if f is a continuous function on [a,b] and f(a)<M<f(b), there exists some c∈[a,b] such that f(c)=M".

Now, we will apply the theorem for the given function f(x).

So,

f(0) = 0^2 +7(0) + 1 = 1

And,

f(9)=9² + 7(9) + 1 = 81 + 63 + 1 = 145

Here,  f(0) = 1< 19< 145 = f(9).

So, f is continous since it is a polynomial. Then the IVT applies, and such c exists.

To find, c,

We have to solve the quadratic equation f(c) =19.

This equation is

c² + 7c + 1 = 19.

Rearranging, c²+ 7c - 18=0.

Factor the expression to get

c² + 9c - 2c -18 = 0

⇒ c(c + 9) - 2( c + 9) = 0

⇒ (c - 2)(c + 9) = 0

⇒ c = 2 or -9

c = -9 is not possible beacuse it is not in the interval [0, 9].

So, the value of c is 2.

⇒ f(2) = 2^2 + 7(2) + 1 = 4 + 14 + 1 = 19

Hence, we verified the intermidiate value theorem applies to the function f(x) = x^2 + 7x  + 1 . And the value of c is 2.

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