Critical value is the value of an independent variable corresponding to a critical point (Function will not be differentiable or its derivative is zero) in the function. It is a point where if the value goes above or below it will lead to significant change in other values also. Critical value plays an important role in calculus and statistics.
Critical Value

Critical Value in Calculus
In a function say f(x), critical point is the value in the domain of f where the function will not be differentiable or the derivative leads to zero. Critical points can either have a maxima or minima. It can easily found by differentiating the given function and then solving for f(x) = 0. If the derivative is zero the point will be considered as stationary point in the function.

Critical Value in Statistics
If the value of the test statistic is greater than the critical value then we reject the null hypothesis and accept alternative hypothesis and if the value for test statistic is less than the critical value alternative hypothesis will be accepted. It is the value of a test variable between acceptance and rejection. In statistics for hypothesis testing we make use of critical values.
Critical value tables can be found for different tests like z test, t test, f test, Mann Whitney test etc.,
Given below is the critical value table for z test
Z
0.00
0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09
0.0 0.0000 0.0040 0.0080 0.0120 0.0160 0.0199 0.0239
0.0279 0.0319 0.0359
0.1 0.0398 0.0438 0.0478 0.0517 0.0557 0.0596 0.0636
0.0675
0.0714 0.0753
0.2 0.0793 0.0832
0.0871 0.0910 0.0948 0.0987 0.1026
0.1064
0.1103 0.1141
0.3 0.1179 0.1217 0.1255 0.1293 0.1331 0.1368 0.1406
0.1443
0.1480 0.1517
0.4 0.1554 0.1591 0.1628 0.1664 0.1700 0.1736 0.1772
0.1808
0.1844 0.1879
0.5 0.1915 0.1950 0.1985 0.2019 0.2054 0.2088 0.2123 0.2157 0.2190 0.2224
0.6 0.2257 0.2291 0.2324 0.2357 0.2389 0.2422 0.2454
0.2486
0.2517 0.2549
0.7 0.2580 0.2611 0.2642 0.2673 0.2704 0.2734 0.2764 0.2794 0.2823 0.2852
0.8 0.2881 0.2910 0.2939 0.2967 0.2995 0.3023 0.3051
0.3078 0.3106 0.3133
0.9 0.3159 0.3186 0.3212 0.3238 0.3264 0.3289 0.3315
0.3340 0.3365 0.3389
1.0 0.3413 0.3438 0.3461 0.3485 0.3508 0.3531 0.3554
0.3577 0.3599 0.3621

The steps for critical value method is given below:
  1. Find the derivative of the given function f(x).
  2. After differentiating the given function equate f'(x) to zero. The solution to the equation f'(x) = 0 will give the critical points.
  3. Critical points can be a maxima or minima or a point of inflexion.
  4. If the function is changing its behavior ( Increasing to decreasing) then it is said to be local maxima else local minima.
  5. In the point of inflexion the behavior does not change the function will either be constantly increasing or decreasing.
Let see with the help of example how to find the critical value for the function, let f(x) = $7x^{3}$ -12 $x^{2}$ + 5 =0

$7x^{3}$ -12 $x^{2}$ + 5 = 0
f '(x) = $21x^{2}$ - 24 x = 0
x(21 x - 24) = 0
x = 0 or x = $\frac{24}{21}$ are the critical values.

Putting x = 0 in the given equation gives
f(0) = 0 - 0 + 5
f(0) = 5
Putting x = $\frac{24}{21}$ in the given equation gives

f($\frac{24}{21}$) = 7($\frac{24}{21}$)$ ^{3}$ - 12($\frac{24}{21}$ )$^{2}$ + 5 = 0

= 10.37 - 15.67 + 5 = 0
= -0.3
In hypothesis testing it is necessary to have these steps to solve the problem.
  1. State the null and alternative hypothesis.
  2. Assuming the null hypothesis to be true give the test statistic for the given data.
  3. For a given level of significance find the critical value using the appropriate tables.
  4. Plug in the known values in the test statistic and find the calculated test statistic.
  5. State the conclusion by comparing the test statistic to the critical value. If the calculated test statistic value is less than the critical value we accept the null hypothesis and vice-versa.

Solved Examples

Question 1: Find the critical value for the function f(x) = $x^{2}$ + 12x + 20 = 0
Solution:
 
The given equation f(x) =  x$^{2}$ + 12x + 20 = 0
f'(x) = 2x + 12
After equating it to zero we get f'( x )= - 6
Plugging in x = - 6  in the given equation
f(x) = (-6)$^{2}$ + 12 (-6) + 20 = 0
f(x) = -16
Therefore we get -16 as the critical value.
 

Question 2: Find the critical value for the function f(x) = $24x^{3}$ - 5x$^{2}$  -50 = 0
Solution:
 
The given equation f(x) =  $24x^{3}$ - 5x$^{2}$  -50 = 0
f'(x) = $72x^{2}$ - 10x = 0
f'(x) = x( 72x -10) = 0

x = 0 or x = $\frac{10}{72}$ are the critical values.

Putting x = 0 in the given equation gives
f(0) = 0 - 0 - 50
f(0) = - 50

Putting x = $\frac{10}{72}$  in the given equation gives

f($\frac{10}{72}$) = 24($\frac{10}{72}$)$ ^{3}$ - 5( $\frac{10}{72}$ )$^{2}$ - 50 = 0

= 0.0658 - 0.098 -50 = 0
= - 50.0322