Consider the following equation f(x) = x^5 (a) -Find the interval on which f is
ID: 2838302 • Letter: C
Question
Consider the following equation f(x) = x^5
(a) -Find the interval on which f is increasing. (Enter your answer using interval notation.)
-Find the interval on which f is decreasing. (Enter your answer using interval notation.)
(b) Find the local minimum value of f.
(c) Find the inflection point. (x, y) =
Find the interval on which f is concave up. (Enter your answer using interval notation.)
Find the interval on which f is concave down. (Enter your answer using interval notation.)
Consider the following equation (x^2)/(X^2-16) find the following:
(a) Find the vertical and horizontal asymptotes.
x
=
(smaller x-value)
x
=
(larger x-value)
y
=
(b) -Find the interval where the function is increasing. (Enter your answer using interval notation.
-Find the interval where the function is decreasing. (Enter your answer using interval notation.)
(c) -Find the local maximum value.
(d) -Find the interval where the function is concave up. (Enter your answer using interval notation.)
-Find the interval where the function is concave down. (Enter your answer using interval notation.)
x
=
(smaller x-value)
x
=
(larger x-value)
y
=
Explanation / Answer
1.
f(x) = x^5
f'(x) = 5x^4
f''(x) = 20x^3
(a)
for increasing, f'(x) > 0
5x^4 > 0
x > 0............increasing
x < 0...........decreasing
(b)
local minimum value exist at critical point i.e. at f'(x) = 0
for f'(x) = 0
x = 0
Local minimum value is f(0) = 0
(c)
for inflection point, f''(x) = 0
20x^3 = 0
x = 0.................inflection point
for concave up, f''(x) > 0
20x^3 > 0
x > 0............concave up
x < 0............concave down
2.
(a)
f(x) = (x^2)/(x^2-16)
vertical asymptote exist at denominator = 0
x^2 - 16 = 0
x = 4...........larger x-value
x = -4.........smaller x-value
for equal coefficient of highest degree numerator and denominator, horizontal aymptote is the ratio of leading coeffcients i.e.
y = 1............horizontal asymptote
(b)
f'(x) = [(x^2 - 16) * 2x - x^2 * 2x] / (x^2 - 16)^2
f'(x) = [2x^3 - 32x - 2x^3] / (x^2 - 16)^2
f'(x) = -32x / (x^2 - 16)^2
f''(x) = -[(x^2 - 16)*32 - 32x * 2x] / (x^2 - 16)^4
f''(x) = -[32x^2 - 512 - 64x^2] / (x^2 - 16)^4
f''(x) = -[-32x^2 - 512] / (x^2 - 16)^4
f''(x) = [32x^2 + 512] / (x^2 - 16)^4
for increasing function, f'(x) > 0
-32x / (x^2 - 16)^2
x < 0.............increasing
x > 0.............decreasing
(b) local minimum value exist at critical point i.e. f'(x) = 0
f'(x) = 0 gives
x = 0
Thus f(0) = 0............local minimum value
(c)
for concave up, f''(x) > 0
[32x^2 + 512] / (x^2 - 16)^4 > 0
since both expressions are always positive
hence it is always concave up
x = (-infinity, infinity)..............concave up
x = null.......................concave down