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f(b)+ f(a)+ The Intermediate Value Theorem states that if f is a continuous on the interval [a, b], then it takes on any given value between f(a) and f(b) at some point in (a, b). We can use the Intermediate Value Theorem to show that the following equation has a solution: $-3z = \sin(4 - x^2 + 1)$ Step 1: Solve this equation for 0. By subtracting the right hand side on both sides of the equation we get: Now define a function f with the left hand side of the equation: f(x) = Step 2: State the domain of f(x). The domain in interval notation is: Step 3: Find a continuous interval [a, b] of f(x) so that f(a) is positive and f(b) is negative. a = b = The function f(x) is continuous on [a, b]. (No answer given) ? We now conclude by the Intermediate Value Theorem that the function f has a root (crosses the x-axis) in the interval (a, b) so the original equation has a solution. Check

          f(b)+
f(a)+
The Intermediate Value Theorem states that if f is a continuous on the interval [a, b], then it takes on any given value between f(a) and f(b) at some point in (a, b).
We can use the Intermediate Value Theorem to show that the following equation has a solution:
$-3z = \sin(4 - x^2 + 1)$
Step 1: Solve this equation for 0. By subtracting the right hand side on both sides of the equation we get:
Now define a function f with the left hand side of the equation:
f(x) = 
Step 2: State the domain of f(x). The domain in interval notation is:
Step 3: Find a continuous interval [a, b] of f(x) so that f(a) is positive and f(b) is negative.
a = 
b = 
The function f(x) is continuous on [a, b]. (No answer given) ?
We now conclude by the Intermediate Value Theorem that the function f has a root (crosses the x-axis) in the interval (a, b) so the original equation has a solution.
Check

f(b)+
f(a)+
The Intermediate Value Theorem states that if f is a continuous on the interval [a, b], then it takes on any given value between f(a) and f(b) at some point in (a, b).
We can use the Intermediate Value Theorem to show that the following equation has a solution:
-3z = sin(4 - x^2 + 1)
Step 1: Solve this equation for 0. By subtracting the right hand side on both sides of the equation we get:
Now define a function f with the left hand side of the equation:
f(x) =
Step 2: State the domain of f(x). The domain in interval notation is:
Step 3: Find a continuous interval [a, b] of f(x) so that f(a) is positive and f(b) is negative.
a =
b =
The function f(x) is continuous on [a, b]. (No answer given) ?
We now conclude by the Intermediate Value Theorem that the function f has a root (crosses the x-axis) in the interval (a, b) so the original equation has a solution.
Check

Added by Juan Francisco L.

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