Let f(x) be the function 11x^2-6x+4. Then the quotient (f(10+h)-f(10))/h can be simplified to ah+b for: a=? b=? My textbook is really confusing and my professor isn't the best and I have a bunch of problems like these that I'm having a lot of trouble even getting started. If you could help it would be greatly appreciated.

Question

anonymous · Answer

which is wrong but my best effort

campbell_st · Answer

ok... so you know 

$$f(x) = 11x^2 - 6x + 4$$

so 

$$f(x + h) = 11(x + h)^2 - 6(x + h) + 4$$

you need to now simplify this 

$$f(x + h) = 11(x^2 + 2xh + h^2) - 6x - 6h + 4$$

which becomes 

$$f(x + h) = 11x^2 + 22xh + 11h^2 - 6x - 6h  + 4$$

so your problem now becomes 

$$\frac{f(x + h) - f(x)}{h} = \frac{11x^2 + 22xh + 4 -6x - 6h + 4 - 11x^2 + 6x - 4}

{h}$$

now you can simplify this by collecting like terms... 

hope it helps

campbell_st · Answer

oops should read, slight typo

$$\frac{f(x + h) - f(x)}{h} = \frac{11x^2 + 22xh + h^2 - 6x - 6h + 4 - 11x^2 + 6x - 4}{h}$$

anonymous · Answer

Thank you, quick question. why isn't the h^2 11h^2?

campbell_st · Answer

ok... well remember you will divide by h  so the will reduce... to jusy h... and you want 

ah + b

so the simplified version is 

$$\frac{f(x + h) - f(x)}{h} = \frac{22xh +h^2 -6h}{h}$$

so divide every term by h, and substitute x = 10

anonymous · Answer

=10

campbell_st · Answer

yes, the question says f(10 + h) 

so replace x with 10 and then simplify