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(a) Start with f(x)= 3x − 5; y=3x - 5, and switch x and y before rearanging to get y in terms of x again:

y=3x - 5

x=3y - 5

(x+5)/3=y

Therefore f^−1(x)=(x+5)/3

(b) Start...

dy/dx = tan³(x)sec²(x)

Integrate both sides ==> ∫dy= ∫ tan³(x)sec²(x) dx

Use the substitution u=tan(x)

And by different...

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...