Question

File attached, please explain #12 (these are practice problems for my final in calc).....thx

Answer 1

the question is\[f(x)=\int\limits_{0}^{x} \sqrt(t^3+1) dt\]
to find f'(2)
for that we have to find f'(x) , when we have to differentiate an integral just plugin the upper limit in the function multiplied with the differentiation of upperlimit this subtracted by the lower limit plugged in and its product with differentiation of lower limit

Answer 2

got it thx.

Answer 3

can u help me out with #13, i posted another thread for that question

Answer 4

so \[f'(x)=\sqrt(x^3+1)*d/dx(x)-\sqrt(0^3+1)*d/dx(0)\]
so we get \[f'(x)=\sqrt(x^3+1)\]
so f'(2)=sqrt(9)
f'(2)=3

Answer 5

thx, if only all my problems were that easy haha

Answer 6

ok
i'll check 13

Answer 7

thx

Answer 8

to find inflection point of a function f(x) we have to find f''(x) first then in the domain of the function we have to check if f''(x) is greater than 0 or less than it , if let for f''(x)=0 x=a then we have to check the sign of f''(x) in left side of a and right side of a , if it changes then a is an inflection point

Answer 9

our question is f''(x)=x(x+1)(x-2)^2
so x=0,-1 and 2 are the roots
to check if x=-1 is an inflection point
if x=-2 we see f''(x)=positive
if x=-0.5 we see f''(x) is negative , so it's an inflection point
to check x=0
if x=-0.5 then f''(x) is negative and for x=0.5 f''(x) is positive so it's an inflection point
to check x=2
if x=0.5 f''(x) is positive
if x=3 then also f''(x) is positive so it's not an inflection point

Answer 10

so only x=-1 and 0 are inflection point, option c