Slope: The Derivative of a Function at a Point

[morrowcosom]

Homework Statement

We are calculating the slope of the function f(x) = 2/x + 3x at x = -3.
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For the function f(x) = 2/x + 3x, we now know:

f(-3) = -29/3
f(-3+h) = (2/( - 3 + h) - 9 + 3h)
Now evaluate the difference quotient, simplifying as much as possible and cancelling h in the denominator:
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(f(-3+h) - f(-3))/h=
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Homework Equations

The Attempt at a Solution

[(2/(-3+h) -9 +3h + (29/3)]/h
= [(2/(-3+h) -27/3 + 29/3]3h/h (eliminate h, second term limit 3)
= [(2/(-3+h) + 2/3]
= [(6/(3(-3+h)) + 2(-3+h)/3(-3+h)]
= 2h/(3(-3+h))

I am doing independent study on a computer program and it it says my answer is wrong. Where did I go astray?


Still working on latex.

 

[Mark44]

Homework Statement

We are calculating the slope of the function f(x) = 2/x + 3x at x = -3.
--------------------------------------------------------------------------------
For the function f(x) = 2/x + 3x, we now know:

f(-3) = -29/3
f(-3+h) = (2/( - 3 + h) - 9 + 3h)
Now evaluate the difference quotient, simplifying as much as possible and cancelling h in the denominator:
--------------------------------------------------------------------------------
(f(-3+h) - f(-3))/h=
--------------------------------------------------------------------------------

Homework Equations

The Attempt at a Solution

[(2/(-3+h) -9 +3h + (29/3)]/h
= [(2/(-3+h) -27/3 + 29/3]3h/h (eliminate h, second term limit 3)

The step above is incorrect. It looks like you just moved 3h outside the brackets, which is not a valid step. The step above it can be simplified somewhat to
[2/(-3+h) +3h + 2/3]/h

Multiply each of 3h and 2/3 by (-3 + h)/(-3 + h) so that you have a common denominator for all three terms in the brackets. Combine like terms and you will be able to factor out an h that can be canceled with the 1/h outside.

= [(2/(-3+h) + 2/3]
= [(6/(3(-3+h)) + 2(-3+h)/3(-3+h)]
= 2h/(3(-3+h))

I am doing independent study on a computer program and it it says my answer is wrong. Where did I go astray?


Still working on latex.

 

[morrowcosom]

The step above is incorrect. It looks like you just moved 3h outside the brackets, which is not a valid step. The step above it can be simplified somewhat to
[2/(-3+h) +3h + 2/3]/h

Multiply each of 3h and 2/3 by (-3 + h)/(-3 + h) so that you have a common denominator for all three terms in the brackets. Combine like terms and you will be able to factor out an h that can be canceled with the 1/h outside.

From here I went:
=(1/h) [(6/(3(-3+h)) + 9h/(3(-3+h) + -6+2h/(3(-3+h))]

and ended up with
=11/(3(-3+h))
and it still says I am wrong. Where did I mess up?

 

[Mark44]

[2/(-3+h) +3h + 2/3]/h

When you get common denominators for fractions, what you're doing is multiplying each of them by 1, which is always valid to do, and doesn't change the value of the fraction being multiplied. The 1 will take different forms, though.

The first fraction in the brackets above is 2/(-3 + h). You multiplied it by 3/3 (= 1) to get 6/(3(-3 + h)). That is correct.

The second fraction is 3h, or 3h/1, to make it actually look like a fraction. To get a denominator of 3(-3 + h), you need to multiply by 3(-3 + h)/[3(-3 + h)]. What you did was multiply by 3/[3(-3 + h)], which isn't 1, and so isn't the right thing to multiply by.

The third fraction looks OK, but needs another pair of parentheses. You multiplied it by (-3 + h)/(-3 + h) to get (-6 + 2h)/(3(-3 + h)).

Fix the one in the middle and then you'll have all three fractions with the same denominator, so you can combine all of the like terms in the three numerators.

 

[morrowcosom]

[2/(-3+h) +3h + 2/3]/h

The second fraction is 3h, or 3h/1, to make it actually look like a fraction. To get a denominator of 3(-3 + h), you need to multiply by 3(-3 + h)/[3(-3 + h)]. What you did was multiply by 3/[3(-3 + h)], which isn't 1, and so isn't the right thing to multiply by.

So, basically when you are talking about multiplying fractions by one to get a common denominator, it is implied that the denominator is also in the numerator, but the h being in the numerator of the middle fraction screws this up, if not I do not understand the concept. If I do not understand, I need to, could you then enlighten me. You have probably gotten an extra thousand posts off of me alone.

I got the right answer to my problem though.

 

[Mark44]

I don't understand what you're saying here: "it is implied that the denominator is also in the numerator"
??
I would actually do this somewhat differently, just getting a common denominator of (-3 + h), which BTW is the same as h - 3.
$$(1/h)\left(\frac{2}{h - 3} + 3h + 2/3\right)$$
$$=(1/h)\left(\frac{2}{h - 3} + \frac{3h(h - 3)}{h - 3} + \frac{2/3 (h - 3)}{h - 3}\right)$$
$$=(1/h)\left(\frac{2 + 3h(h - 3) + (2/3)(h - 3)}{h - 3}\right)$$

Now expand all the terms in the numerator and combine like terms. All terms remaining in the numerator will have at least one factor of h, which you can factor out to cancel with the 1/h outside the fraction.

 

[morrowcosom]

Why is their not a (h-3) next to the 2 in (2/(h-3))?

 

[Mark44]

Because it already has the right denominator.