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Ok, lets do the question 1 first.
inflection point is an x-value that makes f''(x)=0, so the f'(x) doesn't tell us anything about the inflection points. So you can eliminate C, and it should be a critical point/number. A critical number is when f'(x)=0. When is this the case, when x=1, or x=0?
yes, correct, so your answer is...
yes, the first option.
now, question #2 ...
Thanks an okay
See where x=-2.25 and x=-1.75 are on your graph?
this part of the graph you can take alone (and ignore the rest of the graph) to tell me the approximate slope at x=-2.
Don't they land in the middle of a box? How can I find the approx. slope?
You can just visually tell me, if you refer to your options....
you have no function f(x) given in a form of an equation, i know, but you can take that piece from x=-1.75 to x=-2.25, and tell me what the slope of this whole peace is appaorximately
oops not an option
take a narrower space on the function at about x=-2, and try to picture what a tangent line would look like
I still can't really see it. I'm just doing rise/run to count the slope on the graph is this wrong. I'm not really understanding it
it is correct, but you are taking a too wide interval.... i can't copy your picture..... wanted to show something but my extention for copying it isn't there...
We can go ahead and find the function it you want tho'
Is the smaller approx. 1
yes, yes !
if you go from -1.75 to -2
Oh okay lol idk why that was so hard for me sorry
you can take like -1.9 to -2.1
Thank you for your help though!
Okay but isn't the actual answer closer to +1 not -1
Ohh okay I thought -3/4 for a was the approximation
it is not 1, because the graph is squiized, it is more likely a 4.
yeah, if you were to stretch the graph so that ever unit of x and unit of y get the same space, then the slope is way greater than 1, so the answer is 4, actually...
okay thank you!