The graph of f is shown. Apply Rolle's Theorem and find all values of c such that f '(c) = 0 at some point between the labeled intercepts. (Enter your answers as a comma-separated list.)
f (x) = x^2 + 6x − 7
Repost with the intercepts, please.
Would this help these two points were given on the graph illustrated with this question (-7,0) & (1,0)
thnks for the help
To apply Rolle's Theorem to the function f(x) = x^2 + 6x - 7, we need to check if there exist any points between the labeled intercepts where the derivative equals zero.
1. First, let's find the derivative of f(x). The derivative, denoted as f'(x), represents the slope of the function at any given point.
f'(x) = 2x + 6
2. Next, we need to find the x-values of the intercepts of the graph. By setting f(x) = 0, we can solve for x.
x^2 + 6x - 7 = 0
Using the quadratic formula, we can find the intercepts:
x = (-6 ± √(6^2 - 4(1)(-7))) / (2(1))
x = (-6 ± √(36 + 28)) / 2
x = (-6 ± √64) / 2
x = (-6 ± 8) / 2
Solving for x, we have two intercepts:
x₁ = (-6 + 8) / 2 = 1
x₂ = (-6 - 8) / 2 = -7
3. Now, we check if there exist any values of c between these two intercepts (x₁ and x₂) where f'(c) = 0.
For f'(x) = 0:
2x + 6 = 0
2x = -6
x = -3
As x = -3 lies between x₁ = 1 and x₂ = -7, we found a value of c where f'(c) = 0.
Therefore, the only value of c that satisfies the conditions of Rolle's Theorem is c = -3.
Answer: -3
To apply Rolle's Theorem, we need to check if the given function satisfies the conditions of the theorem. Rolle's Theorem states that if a function is continuous on a closed interval [a, b] and differentiable on the open interval (a, b), and if f(a) = f(b), then there exists a point c in the open interval (a, b) such that f '(c) = 0.
In this case, the function f(x) = x^2 + 6x - 7 is a polynomial function, which is continuous and differentiable everywhere. Now, let's find the intercepts of the graph to identify the interval [a, b]:
To find the x-intercepts, set f(x) = 0:
x^2 + 6x - 7 = 0
Next, we can use the quadratic formula to solve for x:
x = (-b ± √(b^2 - 4ac)) / 2a
In this case, a = 1, b = 6, and c = -7. Substituting these values into the formula:
x = (-6 ± √(6^2 - 4(1)(-7))) / (2(1))
x = (-6 ± √(36 + 28)) / 2
x = (-6 ± √64) / 2
x = (-6 ± 8) / 2
Using both solutions:
For x = (-6 + 8) / 2 = 1, we get the point (1, 0).
For x = (-6 - 8) / 2 = -7, we get the point (-7, 0).
So the x-intercepts are (1, 0) and (-7, 0).
Therefore, the interval [a, b] is [-7, 1]. Now, we need to find the critical points in this interval by checking where f '(x) = 0 or where f '(x) is undefined.
To find f '(x), we differentiate f(x) with respect to x. Differentiating f(x) = x^2 + 6x - 7, we get:
f '(x) = 2x + 6
Setting f '(x) = 0 will give us the values of x where f '(c) = 0:
2x + 6 = 0
2x = -6
x = -3
So, x = -3 is the only critical point in the interval [-7, 1]. Therefore, the value of c satisfying f '(c) = 0 at some point between the labeled intercepts is c = -3.
In summary, the value of c such that f '(c) = 0 at some point between the labeled intercepts is c = -3.