Find an equation of the line tangent

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Homework Help Overview

The problem involves finding the equation of the tangent line to the function h(x) = g(f(x)), where f(x) = |sin x| and g(x) = x^2, within the domain -π ≤ x ≤ π. The original poster seeks to determine the domain and range of h(x) and the tangent line at x = π/4.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to establish that h(x) is (|sin x|)^2 and questions the derivative of this function, specifically whether d/dx (|sin x|)^2 equals (|cos x|)^2. Other participants suggest considering the chain rule for differentiation and discuss the implications of squaring the absolute value function.

Discussion Status

Participants are exploring the differentiation of the absolute value function and its implications for the problem at hand. There is a focus on the correct application of differentiation rules, particularly the chain rule, and the need to consider the behavior of the function across different subdomains.

Contextual Notes

Participants note the complexity introduced by the absolute value function, which may require splitting the analysis into cases based on the sign of sin x. There is also an acknowledgment of the need for clarity regarding the derivative of the absolute value function.

gonzalo12345
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Homework Statement



f(x) = |sinx| for - π ≤ x ≤ π
g(x) = x^2
h(x)= g(f(x))

1. Find domain and range of h(x)
2. Find an equation of the line tangent to the graph of h at the point where x= π/4


Homework Equations





The Attempt at a Solution



It think that h(x) is (|sin x|)^2

so, is domain - π ≤ x ≤ π

here is where I am confused:

if d/dx (sin x) = cos x

then

is d/dx (|sin x|)^2 = (|cos x|)^2 ?

thanks in advance for the help.
 
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gonzalo12345 said:
here is where I am confused:

if d/dx (sin x) = cos x

then

is d/dx (|sin x|)^2 = (|cos x|)^2 ?

thanks in advance for the help.


You can't do that. You should consider using the chain rule.
 
danago said:
You can't do that. You should consider using the chain rule.

is there any rule for the derivate for an absolute value?
 
Squaring makes it easy to answer. |f(x)|^2=f(x)^2. Otherwise you have to split it into subdomains where f(x)>=0 and f(x)<0.
 
|x|= x if x\ge 0, -x is x< 0. Its derivative is 1 if x> 0, -1 if x< 0, not defined for x=0.
 

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