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f'(x)=9x²​+4x, and f''(x)=18x+4 (derivatives) 

f'(x)=0 at x=0 or x=-4/9

when x=0 f''(x)>0 therefore a minimum value, when x=-4/9 f''(x)<0 and thus a maximum value. 

Start by rearranging the inequality - make sure the sign next to the x² term is positive to make it easier: 

2x² -5x - 12< 0

Next step is to factorise this quadratic...

Here, we're using the product rule (and the chain rule for the e^-x): y=(x²+1)(e^-x) dy/dx=(2x)(e^-x)+(x²+1)(-1)(e^-x) Then we simplify to get...

We can see that the function is a sum of three terms so we can deal with each term separately and add them up. The term 8x³ and 5 are relatively straightforward and follow the standard rules fo...

dx=du/6 => (u-5)/6=x So the integral is now (2((u-5)/6)-3)(u^1/2) du/6 Which through simplifying becomes (1/36)(2u-28)(u^1/2)du = (1/36)(2u^3/2 -28u^1/2)du After integrating becomes (1/36)(4(u^5/2)/5 -...