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- In the graph of f (x) below, number lines are used to mark where f (x) is zero/positive/negative/undefined, where f ′(x) is zero/positive/negative/undefined, and where f ′′(x) is zero/positive/negative/undefined. Closed circles on the number lines indicate a zero-value, and open-circles indicate an undefined value.
(a) What do the closed circles on the number line for f (x) correspond to on the graph of f (x)?
Solution: This is where f (x) has x-intercepts (zeroes).
(b) How does each + or − sign on the number line for f (x) relate to the graph?
Solution: The + and − signs indicate where f (x) is positive or negative.
(c) What does the closed circle at x = −1 on the number line for f ′(x) correspond to on the graph of f (x)?
Solution: This is where f (x) has a horizontal tangent line.
(d) What does the open circle at x = 2 on the number line for f ′(x) correspond to on the graph of f (x)? Solution: This is where f (x) has a vertical tangent line, so its slope, and f ′(x), is undefined here. (e) How does each + or − sign on the number line for f ′(x) relate to the graph? Solution: They indicate where f ′(x) is positive or negative. f ′(x) is negative where f (x) is decreasing and positive where f (x) is increasing. (f) What does the open circle at x = 2 on the number line for f ′′(x) correspond to on the graph of f (x)?
Solution: Since f ′(x) is undefined here, so is f ′′(x).
(g) How does each + or − sign on the number line for f ′′(x) relate to the graph? Solution: They indicate where f ′′(x) is positive or negative. f ′′(x) is negative where f (x) is concave down and positive where f (x) is concave up.
- For the graph of f (x) shown below, fill in the number lines for f (x), f ′(x) and f ′′(x), marking closed circles where there is a zero, marking open circles for undefined points, and marking + and − signs on each interval to show positive/negativeness.
- Draw a graph of f (x) that fits the information shown in the number lines.
- This problem investigates the derivative of the absolute value function. Recall that we define the absolute value as: |x| =
x if x ≥ 0 , −x if x < 0.
(a) In the space provided, draw a graph of the function f (x) = |x|.
(b) Using your graph from part (a), and your understanding of the derivative as the rate of change/slope of the tangent line, find the derivative function f ′(x) of the above function f (x) = |x|, for x not equal to 0 (fill in the blanks):
f ′(x) =
if x > 0 , (^) Solution: 1 if x > 0
if x < 0 .Solution: -1 if x < 0 . (c) But what about f ′(0)? It is not so clear from the picture even how to draw a tangent line to the function at the origin. So let’s try to compute f ′(0) by first looking at the corresponding lefthand and righthand limits of the difference quotient. i. Compute limh→ 0 + f^ (0+h h)− f^ (0). Hint: use the piecewise definition of f (x) given above. Solution:
lim h→ 0 +
f (0 + h) − f (0) h
= lim h→ 0 +
|h| − | 0 | h
= lim h→ 0 +
h − 0 h
= lim h→ 0 +^
(since h → 0 +^ means h is positive, so |h| = h). ii. Compute limh→ 0 − f^ (0+h h)− f^ (0). Solution:
lim h→ 0 −
f (0 + h) − f (0) h = lim h→ 0 −
|h| − | 0 | h = lim h→ 0 −
−h − 0 h = lim h→ 0 −^
(since h → 0 −^ means h is negative, so |h| = −h). iii. What do your answers to parts (i) and (ii) tell you about f ′(0)? Please explain.
Solution: Since the righthand limit limh→ 0 + f^ (0+h h)− f^ (0)does NOT equal the lefthand limit limh→ 0 − f^ (0+h h)− f^ (0), the (two-sided) limit limh→ 0 f^ (0+h h)− f^ (0)does not exist. But this (two-sided) limit is f ′(0), so f ′(0) does not exist.
- Using what you’ve learned above, sketch the graph of a continuous function g(x) such that g(x) is not differentiable at x = −1, x = 2, nor x = 3.
- Create a piecewise function where one piece is a quadratic function and the other piece is a linear function which is continuous everywhere but not differentiable at x = 0.
f (x) =
if x ≥ 0 ,
if x < 0. . Solution:
f (x) =
x^2 if x ≥ 0
x if x < 0. .
- Create a piecewise function where one piece is a quadratic function and the other piece is a linear function which is continuous and differentiable everywhere.
f (x) =
if x > 0 ,
if x ≤ 0. . Solution:
f (x) =
(x + 1)^2 − 1 if x ≥ 0
2 x if x < 0. . Note, because d dx
x=
(x + 1)^2 − 1 = 2(0 + 1) = 2 and d dx
x=
2 x = 2, so the right and left difference quotient limits agree at x = 0 and therefore the function is differentiable at x = 0.
