Can I please have answers for the following calculus questions?

(1)

If

f(x)= {x^2 if x<= 3

.......{mx+b if x >3

f(x) continuous at x=3 => 3^2 = 3m +b

differentiable at x=3 => 2(3) = m = 6, so b = -9

(2)

Find the derivative of the function

g(x)=(x^2 -4x-1)e^x

g'(x)= (x^2 -4x -1)e^x + (2x -4)e^x by product rule,

= (x^2 -2x -5)e^x

(3)

let f(x)=(x^2+8x+16)/(6x+24)

= (x + 4)^2/ 6(x + 4) = (x + 4)/6 .. x ≠ -4

f '(5)= 1/6

f ''(5)= 0

(4)

Let f(x)= sin(x^5)

f '(x)= -(5x^4)cos(x^5) ... chain rule

f '(1)= -5.cos(1 radian) = -2.702

(5)

Let f(x)=2e^(xsinx)

f '(x)= 2[sinx -xcosx]e^(xsinx)

(6)

Consider the function

f(x)=2x^3 +18x^2 -162x +5, -9<=x<=4

f '(x) = 6x^2 +36x -162 = 6[x^2 +6x -27] = 6(x +9)(x -3)

SPs are at x = -9 and x = 3 both in domain interval..

f ''(x) = 12x + 36 with f ''(-9) <0 so f(-9) a local maxima and

f ''(3) > 0 a local minima

boundary values are f(3) = -265 a minima and f(4) = -227

Absolute minimum value?

f(3) = -265

Absolute maximum value?

f(-9) = 1463

(7)

Find the x coordinate of the absolute maximum for the function

f(x) =11x -5x.ln(x), x>0

f '(x) = 11 - 5[1 +lnx] = 0 => lnx = 6/5

f ''(x) = -5(1/x) < 0 so local maximum

so x_max = e^(6/5) = 3.320

(8)

Find an equation of the tangent line to the curve y = -1 -2x -3x^2 at (1,-6)

dy/dx = -2 -6x

slope (y = mx + c) = m = -2 -6(1) = -8

so y = -8x +c

-6 = -8 +c => c = 2

y = -8x +2

(9)

Use implicit differentiation to find the slope of the tangent line to the curve:

3x^2 -3xy +3y^3 = -192 at the point (-4,-4)

6x -3xy' -3y +9y'y^2 = 0 but y' = m at given point

6(-4) -3(-4)m -3(-4) +9m(-4)^2 = 0

-24 +12m +12 +9*16m = 0

-2 + m +1 +12m = 0

m = 1/13

2 days ago - 2