# FIND CRITICAL NUMBERS OF A FUNCTION

To find critical numbers, we have to use steps given below.

Step 1 :

Find the first derivative.

Step 2 :

Equate the first derivative to 0.

Step 3 :

Solve for the variable. From this, we can understand that slope will be zero at these positions.

Find any critical numbers of the function.

Problem 1 :

f(x) = x3 - 3x2

Solution :

f(x) = x3 - 3x2

f'(x) = 3x2 - 6x

3x2 - 6x = 0

3x(x - 2) = 0

3x = 0 and x - 2 = 0

x = 0 and x = 2

So, the critical numbers are x = 0 and x = 2.

Problem 2 :

g(x) = x4 - 4x2

Solution :

g(x) = x4 - 4x2

g'(x) = 4x3 - 8x

4x3 - 8x = 0

4x(x2 - 2) = 0

4x = 0 and x2 - 2 = 0

x = 0 and x2 = 2

x = ±√2

So, the critical numbers are x = ±√2.

Problem 3 :

g(t) = t√(4 - t), t < 3

Solution :

So, the critical numbers are t = 8/3.

Problem 4 :

Solution :

(x2 + 1)(4) - (4x)(2x) = 0

4x2 + 4 - 8x2 = 0

-4x2 + 4 = 0

-4(x2 - 1) = 0

x2 = 1

x = ±1

So, the critical numbers are x = 1 or -1.

Problem 5 :

h(x) = sin2x + cos x

0 < x < 2π

Solution :

h(x) = sin2x + cos x

h'(x) = 2 sin x cos x - sin x

h'(x) = 0

2 sin x cos x - sin x = 0

sin x(2 cos x  - 1) = 0

If we solve we get,

sin x = 0 and 2 cos x - 1 = 0

x = 0 and 2 cos x = 1

cos x = 1/2

The cos of the function is positive. General solution for the cosine function

x = 2nπ ± θ where θ ∈ (0, π]

For π/3,

 n = 0x = π/3 n = 1x = 2π + π/3x = 7π/3 ∉ (0, π] n = 1x = 2π - π/3x = 5π/3 ∈ (0, π]

So, the critical numbers is x = π/3, 5π/3.

Problem 6 :

f(θ) = 2 sec θ + tan θ

0 < θ < 2π

Solution :

f(θ) = 2 sec θ + tan θ

f'(θ) = 2 sec θ tan θ + sec2θ

f'(θ) = 0

2 sec θ tan θ + sec2θ = 0

2 sin θ + 1 = 0

sin θ = -1/2

The sin of the function is negative .

So, theta is third and fourth quadrant.

Hence, the interval 0 < θ < 2π.

So, the critical numbers is θ = 7π/6, 11π/6.

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