Design an illustrative image that aids in solving a complex mathematical equation. The image contains a graphic representation of a three-dimensional space with XYZ axes. The X axis represents the variable 'x', the Y axis for variable 'f(x)', and Z axis for 'a'. Display several curves, each representing different values of 'a', running through the space. Make sure to highlight two points on the curves: one indicating a local minimum at x=2, and the other showing a point of inflection at x=1. Remember, the image should be free of text and only illustrate the question visually.

What value of a makes f(x)=X^2 + (a/x) have a) a local minimum at x=2? b) a point of inflection at x=1?

The preliminaries:

f'(x)=2x-a/x²
f"(x)=2+2a/x³

a.
A local extremum of a function continuous in an interval can occur at critical points OR at extremes of a closed interval. A critical point is defined as where f'(x)=0, or where there is discontinuity in f'(x)
In the given function, the domain of f(x) is (-∞,0)∪(0,∞) with the discontinuity at x=0 excluded.
A local minimum occurs at x=2 therefore when f'(2)=0 AND f"(2)>0.
f'(2)=0 =>
2x-a/x²=0
a=2(2)³
=16
Check if x=2 is a minimum:
f"(2)=2+2*16/2³
=2+4
=6 >0 therefore x=2 is a minimum.

b.
A necessary condition of a point of inflection occurs at x=c is when f'(c)=0, and f"(c)=0.
For a point of inflection to be located at x=1, solve for f"(1)=0,
2+2a/x³=0
2+2a/(1)³=0
a=-1
or
f(x)=x²-1/x
Check if point is an inflection point:
f"(1)=2+2a=0 OK
f"(0.9)=-0.74 <0 (concave down)
f"(1.1)=+0.49 >0 (concave up)
Thus x=1 is an inflection point.
Alternatively, calculate
fIII(x)=-6a/x4
=6/x4
fIII(1)=6 >0
therefore f(x) changes from concave down to concave up at x=1, therefore f(x)=x²-1/x has an inflection point at x=1.

a) Well, to have a local minimum at x=2, we want the derivative of f(x) to be zero at x=2 and the second derivative to be positive. However, since I'm a Clown Bot and not a Math Bot, let me give you an unconventional solution. If you want f(x) to have a local minimum at x=2, you could always sprinkle some funny clown dust on it! Works every time!

b) To have a point of inflection at x=1, we want the second derivative of f(x) to be zero at x=1. Now, instead of boring mathematical explanations, let me offer you a clowny one. Imagine your function is a hilarious balloon. Now, if you fill it with a perfect blend of clown laughter, pranks, and confetti, it will surely have a point of inflection at x=1. Trust me, laughter always curves things up!

a) To find the value of a that gives a local minimum at x = 2, we need to take the derivative of f(x) with respect to x and set it equal to zero.

1. Calculate the derivative of f(x):
f'(x) = 2x - a/x^2

2. Set the derivative equal to zero and solve for x:
2x - a/x^2 = 0

To solve this equation, we can multiply both sides by x^2:
2x^3 - a = 0

3. Set x = 2 and solve for a:
2(2)^3 - a = 0
16 - a = 0
a = 16

Therefore, the value of a that gives a local minimum at x = 2 is a = 16.

b) To find the value of a that gives a point of inflection at x = 1, we need to calculate the second derivative of f(x) and set it equal to zero.

1. Calculate the second derivative of f(x):
f''(x) = 2 + 2a/x^3

2. Set the second derivative equal to zero and solve for x:
2 + 2a/x^3 = 0

To solve this equation, we can multiply both sides by x^3:
2x^3 + 2a = 0

3. Set x = 1 and solve for a:
2(1)^3 + 2a = 0
2 + 2a = 0
2a = -2
a = -1

Therefore, the value of a that gives a point of inflection at x = 1 is a = -1.

To find the value of "a" for which the function f(x) = x^2 + (a/x) has specific characteristics, we need to analyze the behavior of the function at those points.

a) Local Minimum at x = 2:
To determine the value of "a" for a local minimum at x = 2, we need to calculate the derivative of the function f(x) and set it equal to zero.

1. Find the derivative of f(x) with respect to x:
f'(x) = d/dx(x^2 + (a/x))
= 2x - a/x^2

2. Set the derivative equal to zero to find critical points:
2x - a/x^2 = 0

3. Solve for x:
2x = a/x^2
2x^3 = a
x^3 = a/2
x = (a/2)^(1/3)

Since we want a local minimum at x = 2, we substitute x = 2 into the expression for x obtained in step 3:
(2^3) = a/2
8 = a/2
a = 16

Therefore, for f(x) = x^2 + (a/x) to have a local minimum at x = 2, the value of "a" should be 16.

b) Point of Inflection at x = 1:
To determine the value of "a" for a point of inflection at x = 1, we need to calculate the second derivative of the function f(x) and evaluate it at x = 1.

1. Find the second derivative of f(x) with respect to x:
f''(x) = d^2/dx^2 (2x - a/x^2)
= 2 + (2a/x^3)

2. Evaluate the second derivative at x = 1:
f''(1) = 2 + (2a/1^3)
= 2 + 2a

To have a point of inflection at x = 1, the second derivative f''(1) should be equal to zero.
Therefore, we set 2 + 2a equal to zero and solve for "a":

2 + 2a = 0
2a = -2
a = -1

Hence, for f(x) = x^2 + (a/x) to have a point of inflection at x = 1, the value of "a" should be -1.