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elena-14-01-66 [18.8K]
3 years ago
9

The function f(x) is graphed on the coordinate plane. What is f(−4) ?

Mathematics
1 answer:
elena-14-01-66 [18.8K]3 years ago
8 0
For this type of problem, plug the -4 into the graph for x. So where is the line at x = -4??
y = 4 where x = -4.
That's the same as saying: "what is the y-value when x = -4?)
f(-4) = y = 4
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Consider the equation below. (If an answer does not exist, enter DNE.) f(x) = x3 − 6x2 − 15x + 4 (a) Find the interval on which
kozerog [31]

Answer:

a) The function, f(x) is increasing at the intervals (x < -1.45) and (x > 3.45)

Written in interval form

(-∞, -1.45) and (3.45, ∞)

- The function, f(x) is decreasing at the interval (-1.45 < x < 3.45)

(-1.45, 3.45)

b) Local minimum value of f(x) = -78.1, occurring at x = 3.45

Local maximum value of f(x) = 10.1, occurring at x = -1.45

c) Inflection point = (x, y) = (1, -16)

Interval where the function is concave up

= (x > 1), written in interval form, (1, ∞)

Interval where the function is concave down

= (x < 1), written in interval form, (-∞, 1)

Step-by-step explanation:

f(x) = x³ - 6x² - 15x + 4

a) Find the interval on which f is increasing.

A function is said to be increasing in any interval where f'(x) > 0

f(x) = x³ - 6x² - 15x + 4

f'(x) = 3x² - 6x - 15

the function is increasing at the points where

f'(x) = 3x² - 6x - 15 > 0

x² - 2x - 5 > 0

(x - 3.45)(x + 1.45) > 0

we then do the inequality check to see which intervals where f'(x) is greater than 0

Function | x < -1.45 | -1.45 < x < 3.45 | x > 3.45

(x - 3.45) | negative | negative | positive

(x + 1.45) | negative | positive | positive

(x - 3.45)(x + 1.45) | +ve | -ve | +ve

So, the function (x - 3.45)(x + 1.45) is positive (+ve) at the intervals (x < -1.45) and (x > 3.45).

Hence, the function, f(x) is increasing at the intervals (x < -1.45) and (x > 3.45)

Find the interval on which f is decreasing.

At the interval where f(x) is decreasing, f'(x) < 0

from above,

f'(x) = 3x² - 6x - 15

the function is decreasing at the points where

f'(x) = 3x² - 6x - 15 < 0

x² - 2x - 5 < 0

(x - 3.45)(x + 1.45) < 0

With the similar inequality check for where f'(x) is less than 0

Function | x < -1.45 | -1.45 < x < 3.45 | x > 3.45

(x - 3.45) | negative | negative | positive

(x + 1.45) | negative | positive | positive

(x - 3.45)(x + 1.45) | +ve | -ve | +ve

Hence, the function, f(x) is decreasing at the intervals (-1.45 < x < 3.45)

b) Find the local minimum and maximum values of f.

For the local maximum and minimum points,

f'(x) = 0

but f"(x) < 0 for a local maximum

And f"(x) > 0 for a local minimum

From (a) above

f'(x) = 3x² - 6x - 15

f'(x) = 3x² - 6x - 15 = 0

(x - 3.45)(x + 1.45) = 0

x = 3.45 or x = -1.45

To now investigate the points that corresponds to a minimum and a maximum point, we need f"(x)

f"(x) = 6x - 6

At x = -1.45,

f"(x) = (6×-1.45) - 6 = -14.7 < 0

Hence, x = -1.45 corresponds to a maximum point

At x = 3.45

f"(x) = (6×3.45) - 6 = 14.7 > 0

Hence, x = 3.45 corresponds to a minimum point.

So, at minimum point, x = 3.45

f(x) = x³ - 6x² - 15x + 4

f(3.45) = 3.45³ - 6(3.45²) - 15(3.45) + 4

= -78.101375 = -78.1

At maximum point, x = -1.45

f(x) = x³ - 6x² - 15x + 4

f(-1.45) = (-1.45)³ - 6(-1.45)² - 15(-1.45) + 4

= 10.086375 = 10.1

c) Find the inflection point.

The inflection point is the point where the curve changes from concave up to concave down and vice versa.

This occurs at the point f"(x) = 0

f(x) = x³ - 6x² - 15x + 4

f'(x) = 3x² - 6x - 15

f"(x) = 6x - 6

At inflection point, f"(x) = 0

f"(x) = 6x - 6 = 0

6x = 6

x = 1

At this point where x = 1, f(x) will be

f(x) = x³ - 6x² - 15x + 4

f(1) = 1³ - 6(1²) - 15(1) + 4 = -16

Hence, the inflection point is at (x, y) = (1, -16)

- Find the interval on which f is concave up.

The curve is said to be concave up when on a given interval, the graph of the function always lies above its tangent lines on that interval. In other words, if you draw a tangent line at any given point, then the graph seems to curve upwards, away from the line.

At the interval where the curve is concave up, f"(x) > 0

f"(x) = 6x - 6 > 0

6x > 6

x > 1

- Find the interval on which f is concave down.

A curve/function is said to be concave down on an interval if, on that interval, the graph of the function always lies below its tangent lines on that interval. That is the graph seems to curve downwards, away from its tangent line at any given point.

At the interval where the curve is concave down, f"(x) < 0

f"(x) = 6x - 6 < 0

6x < 6

x < 1

Hope this Helps!!!

5 0
3 years ago
A circle has a circumference of
d1i1m1o1n [39]

Answer:

the diameter is 1

Step-by-step explanation:

Since C = pi*d, solving for d would give 1

5 0
3 years ago
A board is 85cm in length and must be cut so that one piece is 11cm longer than the other piece . Find the length of each piece.
iVinArrow [24]

The shorter piece is 37 cm

The longer piece is 48 cm:

Step-by-step explanation:

Divide 85 cm by 2 = 42.5

Divide 11 cm by 2 = 5.5

Add them together = 48 cm

85 cm - 48 cm = 37 cm

8 0
3 years ago
MATH HELP PLEASE !!!! Tickets to a local movie were sold at $5.00 for adults and $3.00 for students. If 150 tickets were sold fo
Vaselesa [24]

Answer:

A = 80   S =70

Step-by-step explanation:

#Adults tickets = A

#Students tickets = S

A + S = 150

5A (means the price is $5 times the number of Adult tickets) equals the total amount for all of the Adult tickets altogether

3S (means the price is $3 times the number of Student tickets) equals the total amount for all of the students tickets altogether

5A + 3S = 610    (Means adding the total of all the student tickets plus adult tickets will equal $610 for all tickets sold)

Using both equations now, you can use substitution or elimination to solve for one of the variables. Then you can use the variable to substitute to solve the remaining one.

A + S = 150

5A + 3S = 610

Substitution:

A = 150 - S  (Rearrange the first equation by moving S to the other side)

Substitute into the other equation

5 (150 - S) + 3S = 610

750 - 5S + 3S = 610      Combine like terms :    -2S = -140

Solve for S = 70

Substitute into A + S = 150        A + (70) = 150

A = 80

3 0
3 years ago
A parachutist's speed during a free fall reaches 75 meters per second. What is this speed in feet per second? At this speed, how
Alex

<span>Polly uses a probability simulator to pull colored ribbons from a bag 60 times and to flip a coin 60 times. The results are shown in the tables below:

</span>

<span><span>Color of RibbonNumber of Times Pulled</span><span><span>White25</span><span>Green14</span><span>Orange21</span></span></span>

<span><span>HeadsTails</span><span>3822</span></span>

<span>Using Polly's simulation, what is the probability of pulling an orange ribbon and the coin landing on tails?</span>
3 0
3 years ago
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