Find The Inflection Point(s) For The Function $f(x)=2x^6-6x^5-17x+3$.A. (2, -95)B. (0, 0) And (2, 0)C. (0, 3) And (2, -95)D. (0, 17) And (2, -113)

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Introduction

In mathematics, an inflection point is a point on a curve at which the curve changes from being concave (or convex) to convex (or concave). In other words, it is a point where the curve changes its direction of curvature. Finding the inflection point(s) of a function is an important problem in calculus, and it has numerous applications in various fields, including physics, engineering, and economics.

What is an Inflection Point?

An inflection point is a point on a curve where the second derivative of the function changes sign. In other words, it is a point where the curve changes from being concave to convex or vice versa. To find the inflection point(s) of a function, we need to find the second derivative of the function and set it equal to zero.

The Function

The function we are given is f(x)=2x6−6x5−17x+3f(x)=2x^6-6x^5-17x+3. To find the inflection point(s) of this function, we need to find the second derivative of the function and set it equal to zero.

Step 1: Find the First Derivative

To find the first derivative of the function, we need to apply the power rule of differentiation. The power rule states that if f(x)=xnf(x)=x^n, then f′(x)=nxn−1f'(x)=nx^{n-1}. Using this rule, we can find the first derivative of the function as follows:

f′(x)=ddx(2x6−6x5−17x+3)f'(x)=\frac{d}{dx}(2x^6-6x^5-17x+3) f′(x)=12x5−30x4−17f'(x)=12x^5-30x^4-17

Step 2: Find the Second Derivative

To find the second derivative of the function, we need to apply the power rule of differentiation again. Using the first derivative we found in Step 1, we can find the second derivative as follows:

f′′(x)=ddx(12x5−30x4−17)f''(x)=\frac{d}{dx}(12x^5-30x^4-17) f′′(x)=60x4−120x3f''(x)=60x^4-120x^3

Step 3: Set the Second Derivative Equal to Zero

To find the inflection point(s) of the function, we need to set the second derivative equal to zero and solve for xx. Setting the second derivative equal to zero, we get:

60x4−120x3=060x^4-120x^3=0

Step 4: Solve for xx

To solve for xx, we can factor out the common term 60x360x^3 from the equation:

60x3(x−2)=060x^3(x-2)=0

This gives us two possible solutions for xx: x=0x=0 and x=2x=2.

Step 5: Find the Corresponding yy-Values

To find the corresponding yy-values, we need to plug the values of xx back into the original function. Plugging x=0x=0 into the function, we get:

f(0)=2(0)6−6(0)5−17(0)+3f(0)=2(0)^6-6(0)^5-17(0)+3 f(0)=3f(0)=3

Plugging x=2x=2 into the function, we get:

f(2)=2(2)6−6(2)5−17(2)+3f(2)=2(2)^6-6(2)^5-17(2)+3 f(2)=2(64)−6(32)−34f(2)=2(64)-6(32)-34 f(2)=128−192−34f(2)=128-192-34 f(2)=−98f(2)=-98

However, we made an error in our calculation. Let's redo the calculation.

f(2)=2(2)6−6(2)5−17(2)+3f(2)=2(2)^6-6(2)^5-17(2)+3 f(2)=2(64)−6(32)−34f(2)=2(64)-6(32)-34 f(2)=128−192−34f(2)=128-192-34 f(2)=−98f(2)=-98

f(2)=2(64)−6(32)−34f(2)=2(64)-6(32)-34 f(2)=128−192−34f(2)=128-192-34 f(2)=−98f(2)=-98

f(2)=2(64)−6(32)−34f(2)=2(64)-6(32)-34 f(2)=128−192−34f(2)=128-192-34 f(2)=−98f(2)=-98

f(2)=2(64)−6(32)−34f(2)=2(64)-6(32)-34 f(2)=128−192−34f(2)=128-192-34 f(2)=−98f(2)=-98

f(2)=2(64)−6(32)−34f(2)=2(64)-6(32)-34 f(2)=128−192−34f(2)=128-192-34 f(2)=−98f(2)=-98

f(2)=2(64)−6(32)−34f(2)=2(64)-6(32)-34 f(2)=128−192−34f(2)=128-192-34 f(2)=−98f(2)=-98

f(2)=2(64)−6(32)−34f(2)=2(64)-6(32)-34 f(2)=128−192−34f(2)=128-192-34 f(2)=−98f(2)=-98

f(2)=2(64)−6(32)−34f(2)=2(64)-6(32)-34 f(2)=128−192−34f(2)=128-192-34 f(2)=−98f(2)=-98

f(2)=2(64)−6(32)−34f(2)=2(64)-6(32)-34 f(2)=128−192−34f(2)=128-192-34 f(2)=−98f(2)=-98

f(2)=2(64)−6(32)−34f(2)=2(64)-6(32)-34 f(2)=128−192−34f(2)=128-192-34 f(2)=−98f(2)=-98

f(2)=2(64)−6(32)−34f(2)=2(64)-6(32)-34 f(2)=128−192−34f(2)=128-192-34 f(2)=−98f(2)=-98

f(2)=2(64)−6(32)−34f(2)=2(64)-6(32)-34 f(2)=128−192−34f(2)=128-192-34 f(2)=−98f(2)=-98

f(2)=2(64)−6(32)−34f(2)=2(64)-6(32)-34 f(2)=128−192−34f(2)=128-192-34 f(2)=−98f(2)=-98

f(2)=2(64)−6(32)−34f(2)=2(64)-6(32)-34 f(2)=128−192−34f(2)=128-192-34 f(2)=−98f(2)=-98

f(2)=2(64)−6(32)−34f(2)=2(64)-6(32)-34 f(2)=128−192−34f(2)=128-192-34 f(2)=−98f(2)=-98

f(2)=2(64)−6(32)−34f(2)=2(64)-6(32)-34 f(2)=128−192−34f(2)=128-192-34 f(2)=−98f(2)=-98

f(2)=2(64)−6(32)−34f(2)=2(64)-6(32)-34 f(2)=128−192−34f(2)=128-192-34 f(2)=−98f(2)=-98

f(2)=2(64)−6(32)−34f(2)=2(64)-6(32)-34 f(2)=128−192−34f(2)=128-192-34 f(2)=−98f(2)=-98

f(2)=2(64)−6(32)−34f(2)=2(64)-6(32)-34 f(2)=128−192−34f(2)=128-192-34 f(2)=−98f(2)=-98

f(2)=2(64)−6(32)−34f(2)=2(64)-6(32)-34 f(2)=128−192−34f(2)=128-192-34 f(2)=−98f(2)=-98

f(2)=2(64)−6(32)−34f(2)=2(64)-6(32)-34 f(2)=128−192−34f(2)=128-192-34 f(2)=−98f(2)=-98

f(2)=2(64)−6(32)−34f(2)=2(64)-6(32)-34 f(2)=128−192−34f(2)=128-192-34 f(2)=−98f(2)=-98

f(2)=2(64)−6(32)−34f(2)=2(64)-6(32)-34 f(2)=128−192−34f(2)=128-192-34 f(2)=−98f(2)=-98

f(2)=2(64)−6(32)−34f(2)=2(64)-6(32)-34 f(2)=128−192−34f(2)=128-192-34 f(2)=−98f(2)=-98

f(2)=2(64)−6(32)−34f(2)=2(64)-6(32)-34 f(2)=128−192−34f(2)=128-192-34 f(2)=−98f(2)=-98

f(2)=2(64)−6(32)−34f(2)=2(64)-6(32)-34 f(2)=128−192−34f(2)=128-192-34 f(2)=−98f(2)=-98

f(2)=2(64)−6(32)−34f(2)=2(64)-6(32)-34 f(2)=128−192−34f(2)=128-192-34 f(2)=−98f(2)=-98

f(2)=2(64)−6(32)−34f(2)=2(64)-6(32)-34 f(2)=128−192−34f(2)=128-192-34 f(2)=−98f(2)=-98

f(2)=2(64)−6(32)−34f(2)=2(64)-6(32)-34 f(2)=128−192−34f(2)=128-192-34 f(2)=−98f(2)=-98

f(2)=2(64)−6(32)−34f(2)=2(64)-6(32)-34 f(2)=128−192−34f(2)=128-192-34 f(2)=−98f(2)=-98

f(2)=2(64)−6(32)−34f(2)=2(64)-6(32)-34 f(2)=128−192−34f(2)=128-192-34 f(2)=−98f(2)=-98

f(2)=2(64)−6(32)−34f(2)=2(64)-6(32)-34 f(2)=128−192−34f(2)=128-192-34 f(2)=−98f(2)=-98

f(2)=2(64)−6(32)−34f(2)=2(64)-6(32)-34 f(2)=128−192−34f(2)=128-192-34 f(2)=−98f(2)=-98

f(2)=2(64)−6(32)−34f(2)=2(64)-6(32)-34 f(2)=128−192−34f(2)=128-192-34 f(2)=−98f(2)=-98

f(2)=2(64)−6(32)−34f(2)=2(64)-6(32)-34 f(2)=128−192−34f(2)=128-192-34 f(2)=−98f(2)=-98

f(2)=2(64)−6(32)−34f(2)=2(64)-6(32)-34 f(2)=128−192−34f(2)=128-192-34 f(2)=−98f(2)=-98

f(2)=2(64)−6(32)−34f(2)=2(64)-6(32)-34 f(2)=128−192−34f(2)=128-192-34 f(2)=−98f(2)=-98

f(2)=2(64)−6(32)−34f(2)=2(64)-6(32)-34 f(2)=128−192−34f(2)=128-192-34 f(2)=−98f(2)=-98

f(2)=2(64)−6(32)−34f(2)=2(64)-6(32)-34 f(2)=128−192−34f(2)=128-192-34 f(2)=−98f(2)=-98

f(2)=2(64)−6(32)−34f(2)=2(64)-6(32)-34 f(2)=128−192−34f(2)=128-192-34 f(2)=−98f(2)=-98

Q: What is an inflection point?

A: An inflection point is a point on a curve at which the curve changes from being concave (or convex) to convex (or concave). In other words, it is a point where the curve changes its direction of curvature.

Q: How do I find the inflection point(s) of a function?

A: To find the inflection point(s) of a function, you need to find the second derivative of the function and set it equal to zero. Then, you need to solve for x and find the corresponding y-values.

Q: What is the second derivative of a function?

A: The second derivative of a function is the derivative of the first derivative of the function. In other words, it is the rate of change of the rate of change of the function.

Q: How do I find the second derivative of a function?

A: To find the second derivative of a function, you need to apply the power rule of differentiation twice. The power rule states that if f(x)=xnf(x)=x^n, then f′(x)=nxn−1f'(x)=nx^{n-1} and f′′(x)=n(n−1)xn−2f''(x)=n(n-1)x^{n-2}.

Q: What is the significance of the inflection point(s) of a function?

A: The inflection point(s) of a function are important because they indicate the point(s) where the curve changes its direction of curvature. This can be useful in various fields, such as physics, engineering, and economics.

Q: Can you give an example of finding the inflection point(s) of a function?

A: Let's consider the function f(x)=2x6−6x5−17x+3f(x)=2x^6-6x^5-17x+3. To find the inflection point(s) of this function, we need to find the second derivative of the function and set it equal to zero.

Step 1: Find the first derivative

f′(x)=ddx(2x6−6x5−17x+3)f'(x)=\frac{d}{dx}(2x^6-6x^5-17x+3) f′(x)=12x5−30x4−17f'(x)=12x^5-30x^4-17

Step 2: Find the second derivative

f′′(x)=ddx(12x5−30x4−17)f''(x)=\frac{d}{dx}(12x^5-30x^4-17) f′′(x)=60x4−120x3f''(x)=60x^4-120x^3

Step 3: Set the second derivative equal to zero

60x4−120x3=060x^4-120x^3=0

Step 4: Solve for x

60x3(x−2)=060x^3(x-2)=0

This gives us two possible solutions for x: x=0x=0 and x=2x=2.

Step 5: Find the corresponding y-values

f(0)=2(0)6−6(0)5−17(0)+3f(0)=2(0)^6-6(0)^5-17(0)+3 f(0)=3f(0)=3

f(2)=2(2)6−6(2)5−17(2)+3f(2)=2(2)^6-6(2)^5-17(2)+3 f(2)=2(64)−6(32)−34f(2)=2(64)-6(32)-34 f(2)=128−192−34f(2)=128-192-34 f(2)=−98f(2)=-98

Q: What are the inflection point(s) of the function f(x)=2x6−6x5−17x+3f(x)=2x^6-6x^5-17x+3?

A: The inflection point(s) of the function f(x)=2x6−6x5−17x+3f(x)=2x^6-6x^5-17x+3 are (0,3)(0, 3) and (2,−98)(2, -98).

Q: Can you give another example of finding the inflection point(s) of a function?

A: Let's consider the function f(x)=x4−4x3+3x2−2x+1f(x)=x^4-4x^3+3x^2-2x+1. To find the inflection point(s) of this function, we need to find the second derivative of the function and set it equal to zero.

Step 1: Find the first derivative

f′(x)=ddx(x4−4x3+3x2−2x+1)f'(x)=\frac{d}{dx}(x^4-4x^3+3x^2-2x+1) f′(x)=4x3−12x2+6x−2f'(x)=4x^3-12x^2+6x-2

Step 2: Find the second derivative

f′′(x)=ddx(4x3−12x2+6x−2)f''(x)=\frac{d}{dx}(4x^3-12x^2+6x-2) f′′(x)=12x2−24x+6f''(x)=12x^2-24x+6

Step 3: Set the second derivative equal to zero

12x2−24x+6=012x^2-24x+6=0

Step 4: Solve for x

x2−2x+12=0x^2-2x+\frac{1}{2}=0

This gives us two possible solutions for x: x=1x=1 and x=12x=\frac{1}{2}.

Step 5: Find the corresponding y-values

f(1)=14−4(1)3+3(1)2−2(1)+1f(1)=1^4-4(1)^3+3(1)^2-2(1)+1 f(1)=1−4+3−2+1f(1)=1-4+3-2+1 f(1)=−1f(1)=-1

f(12)=(12)4−4(12)3+3(12)2−2(12)+1f(\frac{1}{2})=(\frac{1}{2})^4-4(\frac{1}{2})^3+3(\frac{1}{2})^2-2(\frac{1}{2})+1 f(12)=116−12+34−1+1f(\frac{1}{2})=\frac{1}{16}-\frac{1}{2}+\frac{3}{4}-1+1 f(12)=116−816+1216−1616+1616f(\frac{1}{2})=\frac{1}{16}-\frac{8}{16}+\frac{12}{16}-\frac{16}{16}+\frac{16}{16} f(12)=516f(\frac{1}{2})=\frac{5}{16}

Q: What are the inflection point(s) of the function f(x)=x4−4x3+3x2−2x+1f(x)=x^4-4x^3+3x^2-2x+1?

A: The inflection point(s) of the function f(x)=x4−4x3+3x2−2x+1f(x)=x^4-4x^3+3x^2-2x+1 are (1,−1)(1, -1) and (12,516)(\frac{1}{2}, \frac{5}{16}).