1)

So that there is continuity, the middle branch of the function, f(x) = ax +b, must connect with the first and third branches of the function.

So get the point where the 1st branch ends, and the point where the 3rd branch begins, then solve for the middle branch using those two points.

The end of the 1st branch is at x = -1, or, since the 1st branch is not defined at x = -1, the 1st branch tends to finish at x = -1.

So,

f(x), as x --> -1, = 2(-1)^2 -3(-1) +1 = 6.

That's point (-1,6).

The start of the 3rd branch is at x = 2,

So, f(2) = 8(2) +5 = 21

That's point (2,21)

The line that will connect those points has a slope of

m = (21 -6) / (2 -(-1)) = 15/3 = 5

Its equation is, by point-slope form,

y -(6) = 5(x -(-1))

y = 5x +5 +6

y = 5x +11

And that should be the f(x) = ax +b

Hence, a = 5, and b = 11 -----------------answer.

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Lim(x -->0) of tan(4x) / 6x

= 0/0

Use the L'Hopital's rule

= Lim(x --> 0) of sec^2(4x) *4 / 6

= (1)*4 / 6

= 2/3 ----------answer.

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Lim(x --> 0) of [(3x -sin(x))^2] / x^2

= 0/0

Use the L'Hopital's rule.

= Lim(x --> 0) of [2(3x -sin(x))(3 -cos(x)] / 2x

= 0/0

Use the Rule again,

= Lim(x --> 0) of 2[(3x -sin(x))(sin(x)) +(3 -cos(x))(3 -cos(x))] / 2

= 2[0 +(3 -1)(3 -1)] / 2

= 4 -------------------------answer.