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Question: How do you find the y coordinate of the inflection point of the function \(f(x) = 10{(x - 5)^3} + 2?...

How do you find the y coordinate of the inflection point of the function f(x)=10(x−5)3+2?f(x) = 10{(x - 5)^3} + 2?

Explanation

Solution

In this sum, we are asked to find the y coordinate of the inflection point of the given function f(x)=10(x−5)3+2f(x) = 10{(x - 5)^3} + 2 . First, we need to find the second derivative of the given function. Then we have to equate it with zero, that is, the second derivative equals to zero and then find the value of xx . Then we need to substitute it in the original function and find the value of yy.

Complete Step by Step Solution:
Here, we need to find the y coordinate of the inflection point of the given function f(x)=10(x−5)3+2f(x) = 10{(x - 5)^3} + 2 .
We know that the inflection points of a function are the zeroes of the second derivative of the given function.
So we need to find the second derivative first.
The given function is f(x)=10(x−5)3+2f(x) = 10{(x - 5)^3} + 2
The first derivative of the given function is:
f(x)=10(x−5)3+2f(x) = 10{(x - 5)^3} + 2
dydx=ddx10(x−5)3+2\dfrac{{dy}}{{dx}} = \dfrac{d}{{dx}}10{(x - 5)^3} + 2
Splitting the terms,
dydx=10ddx(x−5)3+ddx2\dfrac{{dy}}{{dx}} = 10\dfrac{d}{{dx}}{(x - 5)^3} + \dfrac{d}{{dx}}2
dydx=y′=10(3)(x−5)2\dfrac{{dy}}{{dx}} = y' = 10(3){(x - 5)^2}
dydx=y′=30(x−5)2\dfrac{{dy}}{{dx}} = y' = 30{(x - 5)^2}
This is the first derivative.
Now, let’s find second derivative
y′=30(x−5)2y' = 30{(x - 5)^2}
y′′=ddx30(x−5)2y'' = \dfrac{d}{{dx}}30{(x - 5)^2}
Derivation of y′y' ,
y′′=30(2)(x−5)y'' = 30(2)(x - 5)
y′′=60(x−5)y'' = 60(x - 5)
This is the second derivative.
Now we have to equate it with zero that is, y′′=0y'' = 0
60(x−5)=060(x - 5) = 0
Multiplying the bracket,
60x−300=060x - 300 = 0
Transferring the number to the other side,
60x=30060x = 300
x=30060x = \dfrac{{300}}{{60}}
x=5x = 5
There is only one zero: x=5  x = 5\;
Now substituting x=5  x = 5\;in the original function f(x)=10(x−5)3+2f(x) = 10{(x - 5)^3} + 2
f(x)=yf(x) = y
f(5)=10(x−5)3+2f(5) = 10{(x - 5)^3} + 2
f(5)=10(5−5)3+2f(5) = 10{(5 - 5)^3} + 2
y=2y = 2

Therefore the inflection point of the given function f(x)=10(x−5)3+2f(x) = 10{(x - 5)^3} + 2isP(5,2)P(5,2).

Note: An inflection point is a point on the graph of a function at which the concavity changes.
Points of inflection can occur where the second derivative is zero. In other words, solve y′′=0y'' = 0 to find the potential inflection points.
Also note that all turning points are stationary points, but not all stationary points are turning points.
A point where the derivative of the function is zero but the derivative does not change sign is known as a point of inflection, or saddle point.