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CALC PLEASE HELP!!!!!

Let f(x) = x4 - 2x³.
(i) Find the domain of f(x).
(ii) Compute f'(x) and f"(x).
(iii) Give the coordinates of the critical points.
(iv) Find the intervals where f(x) is increasing / decreasing.
(v) Give the coordinates of the relative extrema. Classify each extrema as a relative maximum
or a relative minimum.
ALSO
(vi) Find the intervals where f(x) is concave up/ concave down.
(vii) Give the coordinates of any points of inflection.
(viii) Sketch the graph of f(x).

CALC PLEASE HELP!!!!! Let f(x) = x4 - 2x³. (i) Find the domain of f(x). (ii) Compute-example-1
User Aecend
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1 Answer

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(i) The domain of f(x) is all real numbers, since there are no restrictions on the input variable x.

(ii) We have:

f(x) = x^4 - 2x^3

f'(x) = 4x^3 - 6x^2

f''(x) = 12x^2 - 12x

(iii) To find the critical points, we need to solve the equation f'(x) = 0:

4x^3 - 6x^2 = 0

2x^2(2x - 3) = 0

x = 0 or x = 3/2

So the critical points are (0,0) and (3/2, -27/16).

(iv) To determine where f(x) is increasing or decreasing, we need to examine the sign of f'(x) on different intervals. We can make a sign chart for f'(x):

| x | -∞ | 0 | 3/2 | +∞ |

|---------|--------|-------|-------|--------|

| f'(x) | - | 0 | + | + |

From the sign chart, we see that f(x) is decreasing on the interval (-∞, 0) and increasing on the interval (0, 3/2) and (3/2, +∞).

(v) To find the relative extrema, we need to examine the sign of f'(x) around the critical points. We can make a table:

| x | 0- | 0+ | 3/2- | 3/2+ |

|---------|-------|-------|-------|-------|

| f'(x) | - | + | - | + |

| f(x) | 0 | 0 | -27/16| -27/16|

From this table, we see that f(x) has a relative minimum of -27/16 at x = 3/2, and no relative maximum or minimum at x = 0.

(vi) To find the intervals where f(x) is concave up or concave down, we need to examine the sign of f''(x) on different intervals.

We can make a sign chart for f''(x):

| x | -∞ | 0 | 1 | +∞ |

|---------|--------|-------|------|--------|

| f''(x) | + | - | + | + |

From the sign chart, we see that f(x) is concave down on the interval (-∞, 0) and concave up on the intervals (0, 3/2) and (3/2, +∞).

(vii) To find the points of inflection, we need to solve the equation f''(x) = 0:

12x^2 - 12x = 0

12x(x - 1) = 0

x = 0 or x = 1

So the points of inflection are (0,0) and (1, -1).

(viii) To sketch the graph of f(x), we can use the information we have gathered so far.

At x = 0, f(x) has a relative minimum of 0 and is concave down. At x = 3/2, f(x) has a relative minimum of -27/16 and is concave up. The point (1, -1) is a point of inflection.

Based on this information, we can sketch a graph of f(x) that looks like this:

```

|

|

|

|

|

|

|

--------o------------o-------

0 3/2 x-axis

```

The graph is a "U" shape that opens upward, with a relative minimum at (3/2, -27/16) and a point of inflection at (1, -1).

User Marcelo De Aguiar
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8.6k points