Can you tell me which of the following functions is increasing for all values of x for which it is defined...?

Question

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x^3 - x

csc x

-e^-x

|x - 4|

or none of these

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Answer 1

I lst you at x... still in 7th grade algebra here.:) I never really got these....

Answer 2

Do you know how to find derivatives?

If there is no value of x that will result in 0 when plugged into the derivative, then there's no change in direction. Then you only need the sign of the derivative to determine if it constantly increases or decreases.

For example:

The derivative of x^3-x = 3 x^2 - 1

If x = -(1/sqrt(3)) or if x = 1/sqrt(3), the slope will equal zero, so there's two changes in direction. It increases until reaching -1/sqrt(3), decreases between -1/sqrt(3) and 1/sqrt(3), increases after 1/sqrt(3).

The fourth one is kind of tricky because of the absolute value. I just know the graph is a V shape wtih the apex at (0,-4). Since it's V-shaped, it obviously doesn't increase for all functions of x.

The third might look kind of tricky, but the derivative of e^u = e^u du. In this case, (-x) is the u. The derivative of -x is -1, so the derivative of -e^-x is e^-x. The negative exponent just means you're looking at the denominator of a fraction - in other words, the derivative is really 1/(e^x). If x is negative, you flip your fraction again, meaning the slope is e^|x|. -e^-x increases from a very negative number to almost zero.

Answer 3

-e^(-x). Its derivative is e^(-x), which is always positive.

All of the other functions have derivatives that are negative for at least some values of x.

Answer 4

-e^-x
e^x is constantly increasing for all x
e^-x is constantly decreasing for all x
so, -e^-x will be constantly increasing