question: on part b aren't you supposed to do reverse chain rule/u substitution? I believe you went straight to reverse power rule but that is a composite function.
I very rarely show work for u-substitution problems. I do not advise that you also take that approach. You're right that's a composition, but it's simple enough to do in your head if you've done thousands of them :)
If you've established that g'(x) = f(x), then you can say that g(x) is increasing if f(x)>0 or decreasing if f(x)0 anyway.) I hope this helps/makes sense...I didn't actually watch the video to see exactly what you're referencing. :)
Not dumb at all. That would be an interval of inflection, which isn't a thing. There isn't a single point between 1 and 3, inclusive, at which the first derivative changes from increasing to decreasing. At x = 1 the first derivative goes from increasing to constant, at every point between 1 and 3 the derivative remains constant, and at x = 3 the derivative goes from constant to decreasing. But no point where the derivative goes from increasing to decreasing. Hope this helps!
Point of Inflection is essentially when f' goes from inc. to dec. and vice versa. From 0 to 1, there is an increase however, the slope is zero again so it doesnt go from inc to dec.
In part A, how do you know that the integral is from 1 to -5? I understand that it gives you f(1), but how am I supposed to make the connection that 1 is the upper limit of integration?
In some ways the answer is experience, but in another sense the answer is definitely: the Fundamental Theorem of Calculus. In problems where we know a value of the function, say f(1) = 3 in this case, and we're asked for another value of the function, say f(-5) in this case, if we have a representation of the derivative of f we can apply the FTC to find the value. I start out by writing a definite integral that involves x = 1 and x = -5, so the integral from -5 to 1 of f' of x is equal to f(1) - f(-5) and then just rearrange to find what we need. Check out page 6 of Notes 16 for a bit more on it: www.turksmathstuff.com/calc-ab-notes.html
Yes you can leave answers unsimplified; you don't need to simplify arithmetic. (I'm not 100% sure about leaving famous trig values like sin(pi/6), though.) I actually encourage my students to simplify whenever possible, though, because (1) they're pretty good at it and it doesn't take very long and (2) simplifying can let you get a better sense of if your answer is reasonable.
Lmao rip i got 0 points on this
Don't worry bud, we are on the same boat.
didn’t we all 🥺
question: on part b aren't you supposed to do reverse chain rule/u substitution? I believe you went straight to reverse power rule but that is a composite function.
I very rarely show work for u-substitution problems. I do not advise that you also take that approach. You're right that's a composition, but it's simple enough to do in your head if you've done thousands of them :)
Wouldn’t you also have to include “f’(x) is positive” in the justification for part C? Or is that unimportant?
If you've established that g'(x) = f(x), then you can say that g(x) is increasing if f(x)>0 or decreasing if f(x)0 anyway.) I hope this helps/makes sense...I didn't actually watch the video to see exactly what you're referencing. :)
really good video bro
Dumb question, why isnt x from 1 to 3 considered a point of inflection the slope goes from positive to negative on g(x) there?
Not dumb at all. That would be an interval of inflection, which isn't a thing. There isn't a single point between 1 and 3, inclusive, at which the first derivative changes from increasing to decreasing. At x = 1 the first derivative goes from increasing to constant, at every point between 1 and 3 the derivative remains constant, and at x = 3 the derivative goes from constant to decreasing. But no point where the derivative goes from increasing to decreasing. Hope this helps!
Point of Inflection is essentially when f' goes from inc. to dec. and vice versa. From 0 to 1, there is an increase however, the slope is zero again so it doesnt go from inc to dec.
In part A, how do you know that the integral is from 1 to -5? I understand that it gives you f(1), but how am I supposed to make the connection that 1 is the upper limit of integration?
In some ways the answer is experience, but in another sense the answer is definitely: the Fundamental Theorem of Calculus. In problems where we know a value of the function, say f(1) = 3 in this case, and we're asked for another value of the function, say f(-5) in this case, if we have a representation of the derivative of f we can apply the FTC to find the value. I start out by writing a definite integral that involves x = 1 and x = -5, so the integral from -5 to 1 of f' of x is equal to f(1) - f(-5) and then just rearrange to find what we need. Check out page 6 of Notes 16 for a bit more on it: www.turksmathstuff.com/calc-ab-notes.html
Can u always leave unsimplified answers for frq’s? I am studying for the test in May, thanks!
Yes you can leave answers unsimplified; you don't need to simplify arithmetic. (I'm not 100% sure about leaving famous trig values like sin(pi/6), though.) I actually encourage my students to simplify whenever possible, though, because (1) they're pretty good at it and it doesn't take very long and (2) simplifying can let you get a better sense of if your answer is reasonable.
For part d, would it be fine to just say the graph of f'(x) changes from decreasing to increasing at x=4?
Yup, that's perfect! Good luck on the exam!
Thank you so much!
What does the three dots mean?
Hi! It's shorthand for therefore. Here's the wikipedia article: en.wikipedia.org/wiki/Therefore_sign
For part B, couldn't you just put the 2(x-4)^2 into your calculator using MATH 9 (TI-84 Plus) with 6 and 3 as limits of integration?
Number 3 isn't a calculator question in 2019 or, trust me, I'd have done it on a calculator!
thank u!