Question

Wondering about derivatives and parenthesis. For instance, with -5/x^2, if x=1, we know that the final answer will be -whatever, but what if x=-1, would it be -whatever or +whatever (underlying question is - are there parenthesis on x (x)^2 or not. Thanks

Answer 1

So you're given the function \[\Large f(x) = \frac{-5}{x^2}\] and you want to find the derivative of this?

Answer 2

No no. That's a random example. I can easily find the derivative (that is an example of a found one). I mean plugging in x values at the end. Like find the derivative of F(x) at x=1

Answer 3

So after that point, you just replace every x with the x value you want, in this case x = 1

I'm not sure what you're asking in terms of parenthesis though

Answer 4

oh did you mean that x^2 is really (x)^2 ?

Answer 5

Yup. That matters with negatives

Answer 6

So I don't know what it is

Answer 7

when we say x^2, we are squaring ALL of the number. If the number is negative, we include it too

so if x = -1, then

x^2 = (x)^2 = (-1)^2 = +1

since -1 times -1 = +1

Answer 8

technically you could go from x^2 to (-1)^2 without having to write (x)^2 but it doesn't hurt to have it in there

Answer 9

you would NOT do this

x^2 = -1^2 = -1

since again, the negative is part of the squaring process

Answer 10

Alright, so if I was doing (for instance), -3/x^2 and a derivative question says at x=-1, the bottom part would actually still be positive.. Never mind you explained it, from what I got the parenthesis are included. Thanks

Answer 11

yes x^2 will never be negative as long as x is a real number

Answer 12

$$x^2 = x • x$$

$$x = –1$$ means replace x with –1

$$(–1)^2 = –1 • –1 = 1$$

Answer 13

$$\large –3^4 = –(3•3•3•3) = –81$$