Question

How do you find the derivative of:

Answer 1

\[\frac{ x^2-3x-4 }{ x-2 }\]

Answer 2

First simplify the numerator into two factors -> \[(x+1)(x-4)\] Since you can't simplify (cancel by the denominator) you will be forced to leave it in its standard form. so don't simplify it. You will use the QUOTIENT rule. Do you know what the QUOTIENT rule for deriving is? If not I will give it to you.

Answer 3

Before we proceed to use the quotient rule, we will look at the original as TWO functions. One function in the NUMERATOR, and one in the DENOMINATOR. Let the function in the denominator be \[g(x)\] and the function in the numerator be \[f(x)\].

The quotient rule states the following:

\[g(x)\times f \prime (x) - g \prime (x)\times f(x) \] DIVIDED by \[g(x)^{2}\]

Answer 4

Sorry, better representation of the QUOTIENT RULE:

\[\frac{ g(x)×f′(x)−g′(x)×f(x) }{ g(x)^2 }\]

Answer 5

What I am trying to find is the critical numbers of the original function
if f (x)=\[\frac{ x^2-3x-4 }{ x-2 }\]
so by using the quotient rule wouldn't I
f ' (x) =

Answer 6

\[\frac{ (x-2)\frac{ d }{ dx }(x^2-3x-4) - (x^2-3x-4)\frac{ d }{ dx } (x-2)}{ (x-2)^2 }\]

Answer 7

yes, perfect. Now derive where you need to derive (where d/dx is).

Answer 8

then it would be
\[\frac{ (x-2)(2x-3) - (x^2 -3x-4)(1) }{ (x-2)^2 }\]

Answer 9

then it would be
\[\frac{ (2x^2-7x+6)-(x^2-3x-4) }{ (x-2)^2 }\]

Answer 10

now
\[\frac{ x^2-4x+10 }{ (x-2)^2 }\]

Answer 11

is this correct so far?

Answer 12

yes, perfect

Answer 13

then I set the numerator and denominator each to zero and solve to get the critical numbers right?

Answer 14

what do you mean by critical numbers?

If you did it that way.. by solving the numerator and denominator seperatley \[f'(x)\] would be 0. So are you trying to find where the gradient for the function f(g(x)) is 0? Is that what you mean by Critical Values?

Answer 15

this problem has 4 steps
A) find the critical numbers of F (if any)
B) find the open interval(s) on which the function is decreasing or increasing
C) apply first derivative test to identify all relative extrema
D) use graphing utility to confirm your answers

Answer 16

\[\frac{ dy }{ dx }[\frac{ f(x) }{ g(x) }] = 0\] for the critical points. So yes. Solve for both the NUMERATOR and DENOMINATOR by putting them = 0 seperatley

Answer 17

Do I need to use the quadratic equation for the numerator?

Answer 18

N.B! The DENOMINATOR CANNOT EQUAL 0!!!

Answer 19

yes, use the quadratic equation.

Answer 20

so the numerator would be \[3\sqrt{14} or \sqrt{14}\]

Answer 21

The denominator actually can be 0 in the case of solving for critical points. As if the DERIVATIVE does not exit when there is no gradient. So solve for both the numerator and denominator equal to 0.

Answer 22

and the denominator \[\neq2\]

Answer 23

so then the denominator is 2

Answer 24

so you have three critical points.

Answer 25

for b: You need to take values on either side of the critical value and plug it into the derivative function. Judge the signs if it is increasing or decreasing. INCREASING: +
DECREASING: -

So construct a little line and plot the points in order. Then work from the left / right of each point as if it is different intervals.

Answer 26

so the critical points are \[(-\infty,\sqrt{14}),(\sqrt{14},2),(2,3\sqrt{14})(3\sqrt{14},\infty)\]

Answer 27

No, the critical points are simply:

\[3\sqrt{14}, \sqrt{14}, 2\]

Answer 28

Those three points are where the GRADIENT (The DERIVATIVE FUNCTION) = 0. That is the CRITICAL POINT(S). Where the DERIVATIVE/GRADIENT is equal to 0

Answer 29

actually I messed up it is root of 6