Calc-- Theorems

Mean Value Theorem Conditions

continuous on [a,b] and differentiable (no sharp changes to slope) on (a,b)

Mean Value Theorem Conclusion

there exists a value c for a<c<b such that f’(c) = (f(b) - f(a)) / (b - a)

Extreme Value Theorem Conditions

The function must be continuous on a closed interval [a,b].

Extreme Value Theorem Conclusion

If conditions are met then f(x) must attain a max and min value on [a,b]

What is a critical point?

Points where the derivative is 0 or und.

Extreme Value Theorem Sentence Explanation

f(x) is continuous so there exists c where a < c < b, such that f(c) >= f(x) and that f(c) <= f(x) for all x on the interval [a,b]

(there will be a maximum and minimum)

Mean Value Theorem Sentence Explantation

f(x) is differentiable and continuous. Therefore, according to the MVT there exists a value where f’(c) = (f(b)-f(a)/(b-a) on the interval (a,b)

(the avg ROC will equal f’(c) at some point)

Reason for a Relative Max

f’(x) switches from positive to negative

Reason for a Relative Min

f’(x) switches from negative to positive

Candidates Test

to find absolute min or max on interval

1. identify critical points

2. make a table of values using f(x) for when x= end points and critical points

3. draw a conclusion using the output

Candidates Test Conclusion

“The absolute max/ min of f(x) on [a,b] is y-value”