Question

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

Answer 1

We are given a cubic function and we have to find its point of inflation. To find the point of inflation. We have to find its second order derivative. To find the derivative we will use the formula
$(ax+b)^n=n.a(x+b)^{n-1}$
After finding the second order derivative will find the critical point to point at which value of the whole expression will be zero. After finding points we will check at which point the curve is concave before it and convex after it. Point at which the curve is concave for the values before it and convex after it at that point will be the inflection point.

Complete step-by-step answer:
Step1: We are given a function whose value is $f\left( x \right) = 10{\left( {x - 5} \right)^3} + 2$ now we will find its second order derivative to find the derivative we will use the formula:
$(ax+b)^n=n.a(x+b)^{n-1}$
$\Rightarrow f'\left( x \right) = 3{\left( {x - 5} \right)^2}$
Again we will differentiate to find the second order derivative:
$\Rightarrow f''\left( x \right) = 6\left( {x - 5} \right)$
Step2: Now we will find the critical points, these are points where f’’(x) =0 or it is undefined. We will find it by equating f”(x) to zero.
$\Rightarrow 6\left( {x - 5} \right) = 0$
$\Rightarrow x - 5 = 0$
Solving for x we will get:
$\Rightarrow x = 5$
Step3: Now we will analyze the concavity:

interval	Test x-value	f”(x)	conclusion
$x < 5$	$x = 4$	$-6$	f is concave down
$x > 5$	$x = 6$	$6$	f is concave up

Step4: Now we will find the inflection point we know that the intervals where f is concave up or down, we can find its inflection points (i.e. where the concavity changes direction).
f is concave down before $x = 5$
f is concave up after $x = 5$
Hence it is the inflection point. Because there is no point at which the function is undefined. So we will consider this only.

Final answer: Hence the inflection point is $x = 5$

Note:
In such types of questions students mainly did not get an approach how to find these points so for this they should find second order derivatives. And then find the critical points. Sometimes students consider critical points as inflection points but this is wrong because every critical point cannot be the inflection point. Because the inflection point is a point before which graph is concave or changes sign and after which graph is convex again changes sign. By using this method we can easily find the point. Sometimes students use graphs to find but using graphs always cannot give us authentic results .Sometimes students don't know how to draw the graphs. So this method is easy and simple which can be used.