Find the critical numbers of a function

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SUMMARY

The critical numbers of the function g(θ) = 28θ − 7 tan θ are determined by finding where the derivative g'(θ) = 28 − 7 sec(θ)^2 equals zero or is undefined. Setting 4 = sec(θ)^2 leads to the solutions θ = ((6n-1)π)/3, where n is any integer. Additionally, the cosine identity indicates that cos(θ) = ±1/2 provides further critical points. Therefore, the complete set of critical numbers includes θ values derived from both sec(θ) and cosine equations.

PREREQUISITES
  • Understanding of calculus, specifically derivatives and critical points.
  • Familiarity with trigonometric functions and identities, particularly secant and cosine.
  • Ability to solve equations involving trigonometric functions.
  • Knowledge of how to express solutions in terms of arbitrary integers.
NEXT STEPS
  • Study the derivation of critical points in trigonometric functions.
  • Learn about the implications of critical numbers in function analysis.
  • Explore the use of trigonometric identities in solving equations.
  • Practice finding critical points for various functions beyond g(θ).
USEFUL FOR

Students studying calculus, particularly those focusing on derivatives and critical points, as well as educators looking for examples of trigonometric function analysis.

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Thank you for viewing my thread. The following is a problem from my homework that is submitted online. I have already exhausted my attempts, but I'm curious as to where I went wrong. I would appreciate if someone could lead me towards the solution.

Homework Statement


Find the critical numbers of the function. (Enter your answers as a comma-separated list. Use n to denote any arbitrary integer values. If an answer does not exist, enter DNE.)
g(θ) = 28θ − 7 tan θ
f2acfb64dbc9f8b72b5f9d2c6e6b696f.png

Homework Equations


The critical values of a function, f, are located where f'(c) = 0 or DNE.

The Attempt at a Solution


g(θ) = 28θ − 7 tan θ
g'(θ) = 28 − 7 sec(θ)^2
0 = 4 - sec(θ)^2
4 = sec(θ)^2, when θ = ((6*n-1)*pi) / 3

74a99371c78723958c43625d14b4e2e6.png
 
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Permanence said:
Thank you for viewing my thread. The following is a problem from my homework that is submitted online. I have already exhausted my attempts, but I'm curious as to where I went wrong. I would appreciate if someone could lead me towards the solution.

Homework Statement


Find the critical numbers of the function. (Enter your answers as a comma-separated list. Use n to denote any arbitrary integer values. If an answer does not exist, enter DNE.)
g(θ) = 28θ − 7 tan θ
f2acfb64dbc9f8b72b5f9d2c6e6b696f.png

Homework Equations


The critical values of a function, f, are located where f'(c) = 0 or DNE.

The Attempt at a Solution


g(θ) = 28θ − 7 tan θ
g'(θ) = 28 − 7 sec(θ)^2
0 = 4 - sec(θ)^2
4 = sec(θ)^2, when θ = ((6*n-1)*pi) / 3

74a99371c78723958c43625d14b4e2e6.png

You have the right idea. Use an identity you know to help you:

##±2 = sec(θ)##
##±2cos(θ) = 1##
##cos(θ) = ±\frac{1}{2}##

Now solve for ##θ##.
 
Permanence said:
Thank you for viewing my thread. The following is a problem from my homework that is submitted online. I have already exhausted my attempts, but I'm curious as to where I went wrong. I would appreciate if someone could lead me towards the solution.

Homework Statement


Find the critical numbers of the function. (Enter your answers as a comma-separated list. Use n to denote any arbitrary integer values. If an answer does not exist, enter DNE.)
g(θ) = 28θ − 7 tan θ
f2acfb64dbc9f8b72b5f9d2c6e6b696f.png

Homework Equations


The critical values of a function, f, are located where f'(c) = 0 or DNE.

The Attempt at a Solution


g(θ) = 28θ − 7 tan θ
g'(θ) = 28 − 7 sec(θ)^2
0 = 4 - sec(θ)^2
4 = sec(θ)^2, when θ = ((6*n-1)*pi) / 3

That is the same as ##2n\pi - \frac \pi 3##. I don't think you have all of them.
 

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