Find an equation of the line tangent

In summary, when given the functions f(x) = |sinx| for -π ≤ x ≤ π, g(x) = x^2, and h(x) = g(f(x)), the domain of h(x) is -π ≤ x ≤ π and the range is [0,1]. To find the equation of the line tangent to the graph of h at x = π/4, the chain rule should be used since d/dx (|sin x|)^2 is not equal to (|cos x|)^2. The derivative of an absolute value function is 1 if x > 0, -1 if x < 0, and undefined at x = 0.
  • #1
gonzalo12345
26
0

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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  • #2
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.
 
  • #3
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?
 
  • #4
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.
 
  • #5
|x|= x if [itex]x\ge 0[/itex], -x is x< 0. Its derivative is 1 if x> 0, -1 if x< 0, not defined for x=0.
 

FAQ: Find an equation of the line tangent

What is the purpose of finding an equation of the line tangent?

The equation of the line tangent is used to represent the slope of a curve at a specific point. This is important in determining the rate of change or gradient of a curve, which can have multiple applications in fields such as physics, engineering, and economics.

How do you find the equation of the line tangent?

The equation of the line tangent can be found by taking the derivative of the given curve and plugging in the x-value of the desired point. This will give the slope of the tangent line, which can then be used to construct the equation of the line using the point-slope form.

Can the equation of the line tangent be used to find the slope at any point on a curve?

Yes, the equation of the line tangent can be used to find the slope at any point on a curve, as long as the curve is differentiable (meaning it has a well-defined tangent line at that point).

What is the relationship between the equation of the line tangent and the derivative?

The equation of the line tangent is directly related to the derivative of a curve. The derivative gives the slope of the tangent line at any point on the curve, and the equation of the line tangent allows us to construct the actual line that represents that slope.

Can the equation of the line tangent be used to find the equation of a curve?

No, the equation of the line tangent only gives information about the slope of a curve at a specific point. To find the equation of a curve, you would need more information such as additional points or the general shape of the curve.

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