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Computing Difference Quotients - Problem 3
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There are actually a couple different kinds of difference quotients. Here is another kind that we haven’t seen yet. F(x) minus f(3) over x minus 3 and you can replace the three by any number it’s still a difference quotient. Let’s calculate that difference quotient for this function; f(x) equals the square root of x plus 1.

F(x) minus f(3) over x minus 3. F(x) is just root x plus 1, f(3) is root 3 plus 1, which is root 4, which is 2. So minus 2, over x minus 3.

When you have a difference quotient that involves a radical function you’re allowed to use the trick of multiplying by the conjugate. The conjugate of this would be root x plus 1 plus 2. You multiply the top and bottom by that. Because remember, when you’re manipulating an expression, you can't just multiply the top by a number. You’ll the change the value of the expression. You’re effectively multiplying by 1, when you multiply the top and the bottom by the same thing.

This becomes a difference of squares. You get root x plus 1 quantity squared, which is x plus 1 minus 2 squared, so minus 4, over x minus 3, times root x plus 1 plus 2. In the numerator, this becomes x minus 3 over x minus 3 times the quantity, root x plus 1, plus 2. And the x minus 3s cancel. You’re left with 1 over root x plus 1 plus 2 and that’s your answer.

When you’re dealing with a difference quotient that involves a radical function, remember this trick of multiplying the top and bottom by the conjugate of the numerator. And this is actually a very desirable form for when you are in calculus and you’re evaluating derivatives using a difference quotient.

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