Question

find the areas of increasing and decreasing and extremum points of the function: y=8X2-Inx

Answer 1

y=x-sinx

Answer 2

can anybody help me solve the above two questions

Answer 3

i can help if you rewrite the first one
is it
\[f(x)=8x^2-\ln(x)\]?

Answer 4

yes

Answer 5

ok so
\[f'(x)=16x-\frac{1}{x}\] and your job is to see where this is positive and negative, that is solve
\[16x-\frac{1}{x}>0\]

Answer 6

f(x)=2x3-15x2+36x-2

Answer 7

\[\frac{16x^2-1}{x}>0\]
\[\frac{16(x+\frac{1}{4})(x-\frac{1}{4})}{x}>0\]

Answer 8

thanks

Answer 9

so you have 4 intervals to consider
\[(\infty, -\frac{1}{4}),(-\frac{1}{4},0),(0,\frac{1}{4}), (\frac{1}{4},\infty)\] and it will be positive on the second and 4th, negative of the first and third, so that gives intervals of increase and decrease respectively

Answer 10

thanks

Answer 11

same idea for
\[f(x)=x-\sin(x)\]
\[f'(x)=1-\cos(x)\]

Answer 12

what of f(x)=2x3-15x2+36x-2

Answer 13

same idea

Answer 14

we can do that next, but notice that
\[1-\cos(x)>0\iff 1>\cos(x)\] so that one is always increasing

Answer 15

ok

Answer 16

\[f(x)=2x^3-15x^2+36x-2\]
\[f'(x)=6x^2-30x+36\] so this one is a quadratic and we can find the zeros probably by factoring

Answer 17

\[x^2-5x+6>0\]
\[(x-3)(x-2)>0\] and since this thing is a parabola that opens up it will be negative between the zeros and positive outside them, so your cubic polynomial will be increasing on
\[(-\infty, 2)\cup (3,\infty)\] and decreasing between

Answer 18

find the velocity and acceleration of the process: x=10e-2t.cos (5t + π/2)