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First we notice that this is a product of two functions of x, so we are going to use the product rule. Recall (uv)'(x)=u'(x)v(x)+v'(x)u(x). Let u(x)=x^(1/2) and v(x)=ln(3x). We need to find u'(x) and v'(x...
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...
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