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!!!
is the final amount after 
years 
is the initial deposit 
is the interest rate in decimal form 
is the number of times the interest is compounded per year
and 
. Since the interest is compounded monthly, it is compounded 12 times per year; therefore, 
. Lets replace those values in our formula:
, 
![2000[(1+ \frac{0.0225}{12} )^{12t}+(1+ \frac{0.03}{12} )^{12t}]=5000](https://tex.z-dn.net/?f=2000%5B%281%2B%20%5Cfrac%7B0.0225%7D%7B12%7D%20%29%5E%7B12t%7D%2B%281%2B%20%5Cfrac%7B0.03%7D%7B12%7D%20%29%5E%7B12t%7D%5D%3D5000%20)
![t[12ln(1.001875)+12ln(1.0025 )]=ln( \frac{5} {2})](https://tex.z-dn.net/?f=t%5B12ln%281.001875%29%2B12ln%281.0025%20%29%5D%3Dln%28%20%5Cfrac%7B5%7D%20%7B2%7D%29)
