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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.
Find dout more information about intermidiate value theorem here:
https://brainly.com/question/14456529
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