Question

Please explain and answer

Answer 1

Point C

`\begin{array}c\cline{1-3}f(x)&f'(x)&f''(x)\\\cline{1-3}A&+&-\\\cline{1-3}B&0&-\\\cline{1-3}C&-&-\\\cline{1-3}D&-&+\\\cline{1-3}E&0&+\\\cline{1-3}F&+&+\\\cline{1-3}\end{array}`

Explanation:

First derivative f'(x)

At a specific point, the slope of a curve is the same as the slope of the tangent line to the curve at that point.

The first derivative, denoted as f'(x), represents the slope of the graph of a function f(x) at a specific point.

A function is increasing when the slope is positive ⇒ f'(x) > 0.

A function is decreasing when the slope is negative ⇒ f'(x) < 0.

A function is stationary when the slope is zero ⇒ f'(x) = 0.

From observation of the provided graph, we can deduce specific points where these conditions apply:

• At points A and F, the slope of the function is positive.
• At points C and D, the slope of the function is negative.
• At points B and E, the slope of the function is zero.

Therefore:

`\begin{array}c\cline{1-3}f(x)&f'(x)&f''(x)\\\cline{1-3}A&+&\\\cline{1-3}B&0&\\\cline{1-3}C&-&\\\cline{1-3}D&-&\\\cline{1-3}E&0&\\\cline{1-3}F&+&\\\cline{1-3}\end{array}`

Second derivative f''(x)

The second derivative, denoted as f''(x), represents the rate of change of the slope of y = f(x).

Concave up curves have increasing slopes as you move from left to right. This corresponds to f''(x) > 0.

Concave down curves have decreasing slopes as you move from left to right. This corresponds to f''(x) < 0.

A point where the curve changes its concavity is called a point of inflection. At a point of inflection, f''(x) = 0.

Stationary points (where f'(x) = 0) can be a local maximum, a local minimum or a stationary point of inflection.

• If f'(x) = 0 and f''(x) > 0, it indicates a local minimum.
• If f'(x) = 0 and f''(x) < 0, it indicates a local maximum.
• If f'(x) = 0 and f''(x) = 0, it could be any one of the three types (maximum, minimum, or stationary point of inflection), so further examination of f''(x) on either side of the stationary point is necessary to determine its exact nature.

From observation of the provided graph:

• Points A and C are located on the curve where it is concave down, so f''(x) < 0 for these points.
• Point B is located at the stationary point (turning point) on the curve where it is concave down, so point B is a local maximum ⇒ f''(x) < 0.
• Points D and F are located on the curve where it is concave up, so f''(x) > 0 for these points.
• Point E is located at the stationary point (turning point) on the curve where it is concave up, so point E is a local minimum ⇒ f''(x) > 0.

Therefore:

`\begin{array}\cline{1-3}f(x)&f'(x)&f''(x)\\\cline{1-3}A&+&-\\\cline{1-3}B&0&-\\\cline{1-3}C&-&-\\\cline{1-3}D&-&+\\\cline{1-3}E&0&+\\\cline{1-3}F&+&+\\\cline{1-3}\end{array}`

So, the point on the graph when y' < 0 and y'' < 0 is point C.

Answer 2

C

Explanation:

did your teacher not explain this ?

f(x) is, well, the function itself.

f'(x) is the change rate of the function at x. that's means it gives us the slope of the tangent at x.

f"(x) is then the change rate of the tangent slopes. are the tangent slopes increasing or decreasing at (and around) x ?

y, y', y" is just a different way to say f(x), f'(x), f"(x).

at A

we see that the function comes from below A and continues to increase further up (towards B).

the function is increasing around A, so the tangent here has a positive slope (from left to right the tangent is pointing upwards), and that means f'(A) > 0.

but we see that the change rate is actually "slowing down" the closer we get to B from A (the function is still growing as mentioned, but at a slower and slower rate). so, the change rate of the change rate (the change rate of the tangent slopes) is negative, and that means f"(A) < 0.

f'(A) = +

f"(A) = -

at B

this is a local maximum (or in general a local extreme point), and the change rate of the function curve is changing its sign exactly at such a point. it goes on this case from increasing to decreasing, and for a moment (exactly this point) it is in balance (0). at any point before that it would be positive, and at any point afterwards it would be negative. so, f'(B) = 0.

the change rate of the change rate, though, was already negative since A (as the increase got smaller and smaller), and this tendency does not stop at B, as the change rate gets smaller and smaller until it reaches B with 0 and it keeps falling below 0 after B. so, the change rate of the change rate is getting even more negative.

so, f"(B) < 0 or = -

at C

the function is decreasing on the way from B to D.

so, the change rate of the function is negative, and f'(C) <0 or = -.

but regarding the change rate of the change rate from B to D the decreasing rate of the function will speed up on its way to D. so, at C, f"(C) < 0 or = -.

at D

the function is still decreasing on the way from C to D (and beyond to E).

so, the change rate of the function is negative, and f'(D) <0 or = -.

regarding the change rate of the change rate : at this point there is no changing of the tendency. the change rate of the tangent slopes is further going up (as the change from further decreasing the slopes to increasing again happened at an unmarked point between C and D), and from D to E the decreasing of the function will further slow down. so, for any point right before and then after D the change rate of the change rate is positive. so, at D, f"(D) > 0.

at E

this is the same situation as at B (just E is a local minimum but still a local extreme point), as the change rate of the function changes its sign again at such a point, and is simply for that very moment 0. so, f'(E) = 0.

the change rate of the change rate, though, was already positive since C (as the decrease got smaller and smaller), and this tendency does not stop at E, as the change rate gets larger and larger until it reaches E with 0 and it keeps climbing above 0 after E. so, the change rate of the change rate is getting even more positive.

so, f"(E) > 0 or = +.

at F

similar to A, but here the change rate of the function speeds up (since E and even further after F). f'(F) > 0 or = +.

the charge rate of the change rate is due to the speeding up also still positive : f"(F) > 0 or = +.