The graph of f is shown Apply Rolles Theorem and find all va

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 + Ax - 21 c =

Solution

Rolles theorem states that if a function y= f(x) is continuous and differentiable in the interval (a,b) such that f(a) = f(b) then there exists some point between a and b where f\'(x) = 0 i.e. its first derivative is = 0;

In this case f(x) =x² + 4x - 21 and a=-7 and b = 3
such that f(-7) = f(3) = 0 (as depicted in the graph)

We are requried to find that point between \'a\' and \'b\' where f\'(x) =0

We have f(x) =x² + 4x - 21
Differentiating it with respect to \'x\' we get:
f\'(x) = 2x+4;
Now Rolle\'s theorem states that f\'(x) will be =0 for some x in hte interval [-7,3]
2x+4 =0
2x=-4
x= -4/2
x=-2

also that x=-2 lies in the interval [-7,3]

Thus, c= -2