CALCULUS EXAM 3 PREP QUESTIONS
AND ANSWERS
If a function is continuous and differentiable on (a,b), then f attains an absolute
maximum and absolute minimum value at some values c and d in (a,b). - ANSWER-
FALSE...the function would need to be continuous on the closed interval [a,b] in order to
attain an absolute maximum and absolute minimum. Consider the function
f (x) = tan x on ⎛ − π/2 , π/2 ⎞. This is continuous on the whole interval, but has no
absolute extrema as there are two vertical asymptotes. The graph tends towards −∞ as
x approaches − π/2 and it tends towards ∞ as x approaches π/2 .
If f has a local maximum or minimum at c, then f '(c) = 0 - ANSWER-FALSE...Consider
the graph of f (x) = |x| . There is a local minimum at 0, but
f '(0) = DNE .
All global extrema are either local extrema or occur at the endpoints of a closed interval.
- ANSWER-True
If f is concave up on an interval, then the graph of f ' is increasing on that interval. -
ANSWER-True
If f (c) exists, f'(c) = 0 , and f''(c) > 0 , then there is a local maximum at x = c . -
ANSWER-FALSE...there is a local minimum at x = c . This comes from the second
derivative test.
In optimization, how do you prove that the critical value is maximum or minimum? -
ANSWER-1. Use the number line or
2. Second derivative test
Find HA: y = (x+2)/(sqrt(x^2+3)) - ANSWER-y=+1 and y=-1
Points of Inflection - ANSWER-If (c, f(c)) is a point of inflection of the graph of f, then
either f"(c)=0 or f" is undefined at c
Critical Value - ANSWER-Where first derivative is 0 or undefined
Find absolute extrema - ANSWER-Use critical values and END POINTS in the function
Horizontal Asymptote Rules - ANSWER-If m>n: NO HA
If m=n: HA = co-eff of m/co-eff of n
If m<n: HA: y = 0
Mean Value Theorem - ANSWER-if f(x) is continuous on [a,b] and differentiable on
(a,b), there is at least one point (x=c) where f'(c)= F(b)-F(a)/b-a
AND ANSWERS
If a function is continuous and differentiable on (a,b), then f attains an absolute
maximum and absolute minimum value at some values c and d in (a,b). - ANSWER-
FALSE...the function would need to be continuous on the closed interval [a,b] in order to
attain an absolute maximum and absolute minimum. Consider the function
f (x) = tan x on ⎛ − π/2 , π/2 ⎞. This is continuous on the whole interval, but has no
absolute extrema as there are two vertical asymptotes. The graph tends towards −∞ as
x approaches − π/2 and it tends towards ∞ as x approaches π/2 .
If f has a local maximum or minimum at c, then f '(c) = 0 - ANSWER-FALSE...Consider
the graph of f (x) = |x| . There is a local minimum at 0, but
f '(0) = DNE .
All global extrema are either local extrema or occur at the endpoints of a closed interval.
- ANSWER-True
If f is concave up on an interval, then the graph of f ' is increasing on that interval. -
ANSWER-True
If f (c) exists, f'(c) = 0 , and f''(c) > 0 , then there is a local maximum at x = c . -
ANSWER-FALSE...there is a local minimum at x = c . This comes from the second
derivative test.
In optimization, how do you prove that the critical value is maximum or minimum? -
ANSWER-1. Use the number line or
2. Second derivative test
Find HA: y = (x+2)/(sqrt(x^2+3)) - ANSWER-y=+1 and y=-1
Points of Inflection - ANSWER-If (c, f(c)) is a point of inflection of the graph of f, then
either f"(c)=0 or f" is undefined at c
Critical Value - ANSWER-Where first derivative is 0 or undefined
Find absolute extrema - ANSWER-Use critical values and END POINTS in the function
Horizontal Asymptote Rules - ANSWER-If m>n: NO HA
If m=n: HA = co-eff of m/co-eff of n
If m<n: HA: y = 0
Mean Value Theorem - ANSWER-if f(x) is continuous on [a,b] and differentiable on
(a,b), there is at least one point (x=c) where f'(c)= F(b)-F(a)/b-a
