Let f(x) = x^2. The following diagram shows part of the...

Let $f(x) = x^2$. The following diagram shows part of the graph of $f$. diagram not to scale The line $L$ is the tangent to the graph of $f$ at the point $A(-k, k^2)$, and intersects the $x$-axis at point $B$. The point $C$ is $(-k, 0)$. The region $R$ is enclosed by $L$, the graph of $f$, and the $x$-axis. This is shown in the following diagram. diagram not to scale a.i. Write down $f'(x)$. a.ii.Find the gradient of $L$. b. Show that the $x$-coordinate of $B$ is $-rac{k}{2}$. c. Find the area of triangle $ABC$, giving your answer in terms of $k$.

Transcript

00:01 Now the first thing i'm just going to double check that that makes sense that they called f of x because the screen is pretty small for me is 8 minus x cubed but if you did the derivative of that you would get negative 3 x squared and what i'm just double checking is that f prime of 1 would be negative 3 because that's what they say for g of x or right they write as y equals so that makes sense then y equals negative 3 x plus 10 and they intersect at the point 1 7 just trying to draw that in there so that makes sense they also give you region r was up here region s is down here and there's one more region but if you're finding the area of the shade of region r what you want so i'll just write area of region r is equal to the integral from 0 to 1 of the upper function which is that negative 3 x plus 10 and you want to subtract off f of x now you want to also make sure you distribute that negative to both terms matter of fact i'll just go ahead and write it because i'm going to rewrite it one more time 8 minus x cubed because while i distribute i also want to write it so i can cancel out i guess i shouldn't say cancel but i can combine like terms at 10 minus 8 is 2 so what i need to do is add 1 to my exponent and then multiply by the reciprocal of my new exponent and what i really like about this is when i plug in 1 and for all those x's i just get the coefficient and plugging in 0 into all those x's will just give me 0 so it's one more thing is to rewrite as 1 4th minus 6 4th plus 8 4th so i have the same denominator and 1 minus 6 is negative 5 plus 8 will give me 3 4th so then in part b we have semi -circle cross sections for that region and this will give us a volume and i don't know if you memorize this formula so i put a little space here that was the answer to a i'm working on b is if you made a semi -circle with cross sections that are the diameter well think about how the diameter is really the base and so the radius is the base divided by 2 and area of one semi -circle would be equal to pi r squared but it's only half of it so divided by 2 so it's pi over 2 times the base over 2 squared as you square both the base and 2 2 squared is 4 times that other 2 would be pi over 8 base squared so this volume is going to you can move that pi over 8 in front and still the integral from 0 to 1 and what's the base well the base is the same…

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