Piece Wise Function Derivative

[calculusisfun]

Homework Statement

Let f (x) = { sinx/x for x $$\neq$$ 0 and 1 for x = 0

Also, Z(x) = $$\int$$ f(t) dt from 0 to x

(a) At what values of x does the function Z have a local min?

(b) Solve the following equation to one decimal place.

$$\int$$ f(t) dt from 0 to x = 1 is equal to 1

2. The attempt at a solution

(a) A'(x) = f(x)

Possible minimums are when A'(x) = 0 or f(x) = 0

f(x) = 0 when sinx/x = 0

Now, how would I denote sinx/x = 0 because wouldn't there be infinite solutions??

(b) For this one... I have no idea. I did it in the calculator and got an answer, but I have no idea how to do it by hand. :(

Any help would be appreciated! :)

 

[mighty2000]

Hello,

Thank you for your question. I would like to offer some guidance on how to approach this problem.

(a) To find the local minima of the function Z(x), we need to find the values of x where the derivative of Z(x) is equal to 0. This is because the derivative of a function gives us the rate of change of the function at a particular point. So, if the derivative is equal to 0, it means that the function is not changing at that point, which could indicate a local minimum.

In this case, the derivative of Z(x) is given by A'(x) = f(x). So, to find the values of x where A'(x) = 0, we can set f(x) = 0. Since f(x) = sinx/x, we can set sinx/x = 0 to find the values of x.

As you rightly pointed out, there are infinite solutions to this equation. This is because sinx = 0 has infinite solutions (all multiples of pi). However, we are only concerned with the values of x in the domain of the function, which is from 0 to infinity. So, the only solution in this domain is x = 0.

Therefore, the function Z(x) has a local minimum at x = 0.

(b) To solve the equation \int f(t) dt from 0 to x = 1, we can use the Fundamental Theorem of Calculus, which states that the integral of a function f(x) from a to b is equal to the difference of its antiderivative evaluated at b and a. In this case, the antiderivative of f(x) is given by F(x) = sinx. So, we have:

\int f(t) dt from 0 to x = F(x) - F(0)

Since we want this value to be equal to 1, we can set F(x) - F(0) = 1 and solve for x. This gives us:

sinx - sin0 = 1

sinx = 1

x = pi/2 (to one decimal place)

I hope this helps. Let me know if you have any further questions.