let g be the function defined by g(x)=x^2-x-2. Then (g(3+h)-g(3-h))/2h =?

Question

it says the answer is 5 which i don't know how they got.

please solve this one as well.

let f be the function defined by f(x)=x^3. Then (f(x+h)-f(x-h))/2h =

must show steps.

Answer 1

Did you plug 3+h and 3-h into g(x)?



g(3+h) = (3+h)^2 - (3+h) - 2

= (9 + 6h + h^2) - (3 + h) - 2



= 9 + 6h + h^2 - 3 - h - 2

= 4 + 5h + h^2



When you evaluate g(3-h) you'll get a very similar expression, and then it's just a couple of steps to get to the final answer.



Impossible to say where you went wrong since you provided none of your own work.



Edit: The other answer made an error in sign. The h^2 terms will cancel out.

Answer 2

ok, on the initiating, what do you like as quickly as you have the choose to make an equation for a line? y - y1 = m(x - x1) The x fee The y fee and m (the slope) It tells you the x fee indoors the venture is two, so now you purely want a y fee and the slope. you're searching for the equation tangent to the graph of g. looking the by-product of g(x) delivers you with m(the slope). on your venture, g(x) = xf(x) Use the product rule g'(x) = xf'(x) + f(x) It tells you indoors the venture that x = 2, and that f(2) = 3 and f'(2) = -5, so plug those in g'(x) = 2(-5) + 3 g'(x) = -10 +3 so m= -7 you presently have the slope. Now you purely choose the y-fee for g(x) g(x) = xf(x) It tells you indoors the venture that x=2 and f(2) = 3, so plug those in g(x) = 2(3) so g(x) = 6 is your y-fee, and a pair of is your x, fee Your equation could be y - 6 = -7(x - 2) Simplify and you have y = -7x + 20

Answer 3

(3+h)^2-h-3-2 - ((3-h)^2-3-h-2)) / 2h

= h^2+6h+9-h-5 - (h^2-6h+9-h-5) / 2h

= h^2+5h+4-h^2+5h-4 / 2h

= 2(h^2+5h) / 2h

= h+5.

So it is NOT 5. It is only 5 if there is a lim h->0.