Question

Let f(x)=6+2x−x^2. Find the difference quotient (f(1+h)−f(1))/h. Simplify your answer.

Answer 1

So let's see if we can identity each piece in the puzzle, then we can just plug then into the difference quotient setup.

\[\large f(\color{orangered}{x})=6+2\color{orangered}{x}-\color{orangered}{x}^2\]
Evaluated at `x=1` gives us,\[\large f(\color{orangered}{1})=6+2(\color{orangered}{1})-(\color{orangered}{1})^2\]

Answer 2

\[\large f(\color{orangered}{1})=7\]
Then we'll also need to evaluate the function at `x=1+h`.\[\large f(\color{orangered}{1+h})=6+2(\color{orangered}{1+h})-(\color{orangered}{1+h})^2\]

Answer 3

Then we just plug these pieces into the `Difference Quotient` form, and simplify.

Let me know if you're confused about how to proceed.
Or about how we plugged in the 1+h.
I know function notation can be a little tricky to get used to.

Answer 4

so this? (3(1+h)^2+2(1+h)+6-(-x^2+2x+6))/h

Answer 5

No... Hmm I'm not really sure what you did there.

Answer 6

\[\large \color{orangered}{f(1)=7}\]\[\large \color{royalblue}{f(1+h)=6+2(1+h)-(1+h)^2}\]

These are the pieces we solved for.
Now plug them into this formula,\[\large \frac{\color{royalblue}{f(1+h)}-\color{orangered}{f(1)}}{h}\]

Answer 7

((6+2(1+h)−(1+h)2)-7)/h?

Answer 8

ya :)

Answer 9

From there, expand out the squared term and simply :D

Answer 10

-1/h

Answer 11

@zepdrix

Answer 12

Hmm I came up with -h^2/h which simplifies to -h.
Lemme check my work again.

Answer 13

Yah still coming up with -h.
Did you remember to distribute the `negative` to each term when you expanded out the square?

Answer 14

ah ok that works thank you very much