OpenStudy (anonymous):

5 years ago
zepdrix (zepdrix):

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,$\large f(\color{orangered}{1})=6+2(\color{orangered}{1})-(\color{orangered}{1})^2$

5 years ago
zepdrix (zepdrix):

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

5 years ago
zepdrix (zepdrix):

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.

5 years ago
OpenStudy (anonymous):

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

5 years ago
zepdrix (zepdrix):

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

5 years ago
zepdrix (zepdrix):

$\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,$\large \frac{\color{royalblue}{f(1+h)}-\color{orangered}{f(1)}}{h}$

5 years ago
OpenStudy (anonymous):

((6+2(1+h)−(1+h)2)-7)/h?

5 years ago
zepdrix (zepdrix):

ya :)

5 years ago
zepdrix (zepdrix):

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

5 years ago
OpenStudy (anonymous):

-1/h

5 years ago
OpenStudy (anonymous):

@zepdrix

5 years ago
zepdrix (zepdrix):

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

5 years ago
zepdrix (zepdrix):

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

5 years ago
OpenStudy (anonymous):

ah ok that works thank you very much

5 years ago