Question

given the conditions for a function (f(x))

f '(x) > 0 on (0,3) | • f '(x) < 0 on (3,∞) |

• f ''(x) > 0 on (0,2) | • f ''(x) < 0 on (−∞,0) ∪ (2,∞) |

how would the graph f(x) look like?

Answer #1

comment if you need further clarification!

For 1 and 2, give a function f that satisfies the given
conditions.
1. f ' (x) = x^5 + 1 + 2 sec x tan x with f(0) = 4
2. f '' (x) = 12x + sin x with f(0) = 3 and f ' (0) = 7

1)Consider the function f(x)f(x) whose second derivative is
f″(x)=9x+8sin(x). If f(0)=4 and f′(0)=2, what is f(5)?
2) Consider the function f(x)=(7/x^3)−(2/x^5).
Let F(x) be the antiderivative of f(x) with F(1)=0.
3)Given that the graph of f(x) passes through the point (5,4)
and that the slope of its tangent line at (x,f(x)) is 3x+3, what
is f(2)?
Then F(2) equals

Let f(x) = x*(2-x) if x>=0, or x*(x+2) if x<0
i) graph the function from x=-3 to x=+3. If you like
WolframAlpha, use
Piecewise[{{x*(2-x),x=>0},{x*(x+2),x<0}}]
If you like Desmos, use f(x)= {x>=0:x*(2-x),
x<0:x*(x+2)}
(for some reason, when you paste that it, it forgets the first
curly-brace { so you’ll need to add it in by hand) Or, you can use
this, but it makes it less clear how to take the derivative:
f(x) = -sign(x)*x*(x - 2*sign(x) )
ii) Find and...

The probability desinity function of X is f(x)=ax+b on the
interval [0,2] and it is 0 elsewhere. You are given that the median
of X is 1.25. Find a and b

2. Show that for the function f : [0, 2] → IR given by f(x) =
x^3 , has f ∈ R[0, 2]

7. (a) Sketch a graph of a function f(x) that satisfies all of
the following conditions.
i. f(2) = 3 and f(1) = −1
ii. lim x→−4 f(x) = −∞
iii. limx→∞ f(x) = 1
iv. lim x→−∞ f(x) = −2
v. lim x→−1+ f(x) = ∞
vi. lim x→−1− f(x) = −∞
vii. f 0 (x) > 0 on (−4, −3.5) ∪ (−2.5, −1.5) ∪ (1, 2) ∪ (2,
∞)
viii. f 0 (x) < 0 on (−∞, −4)...

Sketch the graph of a function f(x) that satisfies all of the
conditions listed below. Be sure to clearly label the axes.
f(x) is continuous and differentiable on its entire domain,
which is (−5,∞)
limx→-5^+ f(x)=∞
limx→∞f(x)=0limx→∞f(x)=0
f(−2)=−4,f′(−2)=0f(−2)=−4,f′(−2)=0
f′′(x)>0f″(x)>0 for −5<x<1−5<x<1
f′′(x)<0f″(x)<0 for x>1x>1

Given the joint probability density function f ( x , y ) for 0
< x < 3 and 0 < y < 2 x^2y/81 Find the conditional
probability distribution of X=1 given that Y = 1 f ( x , y ) = x^2
y/ 81 . F i n d the conditional probability distribution of X=1
given that Y = 1. i . e . f (X ∣ y = 1 )( 1 )

f '(−3) = f '(−2) = f '(0) = f '(1) = 0,
f '(x) < 0 if x < −3 or −2 < x < 0 or x
> 1
f '(x) > 0 if −3 < x < −2 or 0 < x < 1
f ''(x) < 0 if −2.5 < x < −1 or x > 0.5
f ''(x) > 0 if x < −2.5 or −1 < x < 0.5
Sketch the graph of a function

Given the polynomial function f(x)=(x-a)(x-b)^2, and
a<0<b, find the following:
A). Sketch the graph
B). find the x and the y intercepts
C). find the solution to f(x)<0
D). find the solution to f(x) is greater than or equal to 0

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