Inflection Point: Calculus, Graph & Concavity

Ans. Given function: f(x) = x⁴ – 24x² + 11

The first derivative of the function is

f’(x) = 4x³ – 48x

The second derivative of the function is

f”(x) = 12x² – 48

Set f”(x) = 0,

12x² – 48 = 0

Divide by 12 on both sides, we get

x² – 4 = 0

x² = 4

Therefore, x = ± 2

To check or x = 2, substitute x = 1 and 3 in f”(x)

So, f”(1) = 12(1)² – 48 = -36 (negative)

f”(3) = 12(3)² – 48 = 276 (positive)

To check for x = -2, substitute x = 0 and -3 in f”(x)

So, f”(0) = 12(0)² – 48 = -48 (negative)

f”(3) = 12(3)² – 48 = 276 (positive)

Hence proved.

Now, substitute x = ± 2 in f”(x)

Therefore, it becomes

f”(2) = 12(2)² – 48 = -69

f”(-2) = 12(-2)² – 48 = -69

Therefore, the inflection points are (2, -69), and (-2, -69).

Ans. A point of inflection is found where the graph (or image) of a function changes concavity. To find this algebraically, we want to find where the second derivative of the function changes sign, from negative to positive, or vice-versa. So, we find the second derivative of the given function y=x³. 

The first derivative using the power rule

y = xn→y′ = nxn − 1 is,

y′ = 3x² and the seconds derivative is y′′ = 6x. 

We then find where this second derivative equals 0. 6x = 0 when x = 0.

We then look to see if the second derivative changes signs at this point. Both graphically and algebraically, we can see that the function y′′=6x does indeed change sign at, and only at, x = 0, so this is our inflection point.

Ans. In order to find the points of inflection, we need to find f′′(x) using the power rule, f(x) = xn → f′(x) = nxn − 1.

f(x) = −x³ + x² − x + 1

f′(x) = −3x² + 2x − 1

f′′(x) = −6x + 2

Now we set f′′(x) = 0, and solve for x.

−6x + 2=0

−6x = −2

x = 13

To verify this is a true inflection point we need to plug in a value that is less than it and a value that is greater than it into the second derivative. If there is a sign change around the point then it is a true inflection point.

Let x = 0

f′′(0) =−6 (0) + 2 = 2

Now let x = 1

f′′(1) = −6(1) + 2 = −4

Since the sign changes from a positive to a negative around the point x = 13, we can conclude it is an inflection point.

Ans. In order to find the points of inflection, we need to find f′′(t) using the power rule f(x) = xn→f′(x) = nxn − 1.

f(t) = t⁵ −3t³ + 2t +100

f′(t) = 5t⁴ − 9t² + 2

f′′(t) = 20t³ − 18t

Now lets factor f′′(t).

f′′(t) = t(20t²−18)

Now to find the points of inflection, we need to set f′′(t)=0.

t(20t²−18) = 0.

From this equation, we already know one of the point of inflection, t1=0.

To figure out the rest of the points of inflection we can use the quadratic equation.

Recall that the quadratic equation is

t= −b ± b²−4ac−−−−−−−√2a, where a,b,c refer to the coefficients of the equation 

at² + bt + c=0 

In this case, a = 20, b = 0, c = -18.

t = −0 ± 02 − 4(20)(−18)−−−−−−−−−−−−−√(2)(20)

t = ±1440−−−−√40 = ±1210−−√40 = ±310−−√10

Thus the other 2 points of infection are:

t2 = 310−−√10

t3 = −310−−√10

To verify that they are all inflection points we need to plug in values higher and lower than each value and see if the sign changes.

Lets plug in x = −1, x = −12, x = 12, x = 1 

f′′(−1) = 20(−1)³ − 18(−1) = −2

f′′(−12) = 20(−12)³ − 18(−12) = 6.5

f′′(12) = 20(12)³ − 18(12) = −6.5

f′′(1) = 20(1)³ − 18(1) = 2

Since there is a sign change at each point, all are points of inflection.

Ans. In order to find the points of inflection, we need to find f′′(x)

f(x) = x² − 2x + 1

f′(x) = 2x−2

f′′(x) = 2

Now we set f′′(x)=0.

0 = 2.

This last statement says that f′′(x) will never be 0. Thus there are no points of inflection.

Ans. The points of inflection of a given function are the values at which the second derivative of the function is equal to zero.

The first derivative of the function is:

f′(x) = 3x² + 4x +1, and the derivative of this function (the second derivative of the original function), is 

f′′(x) = 6x + 4.

Both derivatives were found using the power rule;

Solving 0 = 6x + 4, x = −23.

To verify that this point is a true inflection point we need to plug in a value that is less than the point and one that is greater than the point into the second derivative. If there is a sign change between the two numbers then the point in question is an inflection point.

Lets plug in x = −1, 

6(−1) + 4 = −2.

Now plug in x = 0

6(0) + 4 = 4.

Therefore, x = −23 is the only point of inflection of the function.

Ans. In order to find all the points of inflection, we first find f′′(x) using the power rule twice f(x) = xn → f′(x) = nxn − 1.

f(x) = x⁴ − x² + 100

f′(x) = 4x³ − 2x

f′′(x) = 12x² − 2

Now we set f′′(x)=0.

12x² − 2 = 0

Thus the points of inflection are x = 1/√6 and x = −1/√6

Ans. To find the inflection point we must find where the second derivative of a function is 0.

Calculating the second derivative is fairly simple. We just need to know that (xn)′=n⋅xn−1:

The first derivative is: 

15x² − 18x + 5 

and if we take the derivative once more we get 

30x − 18.

Setting this equal to zero, we get 

30x − 18 = 0 → x = 18/30 = 3/5

And now all we have to do is plug this value, 3/5, into our original polynomial, to get the answer.

(3/5,−229/25)

Ans. Points of inflection occur when the second derivative changes signs.

y = 3x⁵ − 10x⁴ + 10x³ + 7x + 1

y′ = 15x⁴ − 40x³ + 30x² + 7

y′′ = 60x³ − 120x² + 60x = 60x (x²−2x+1) = 60x(x−1)²

The second derivative equals zero at x = 0 and x = 1. However, the factor (x−1) has degree two in the second derivative. This indicates that x=1 is a root of the second derivative with multiplicity two, so the second derivative does not change signs at this value. It only changes signs at x = 0. Since f(0) = 1, the point of inflection is (0,1).

Ans. The point of inflection of an object can be found by setting the second derivative of the function equal to zero and solving. 

f′(x) = 15x² + 10/3x.

f′′(x) = 30x + 10/3 = 0.

30x = −10/3, x = −1/9.