>>>>>>>>>>>>>>>>> Find the volume of the solid formed by rotating the region enc

Question

>>>>>>>>>>>>>>>>>

Find the volume of the solid formed by rotating the region enclosed by

>>>>>>>>>>>>>>>>>

>>>>>>>>>>>>>>>>>>

>>>>>>>>>>>>>>>>>

>>>>>>>>>>>>>>>>>

Find an equation for the line tangent to the graph of

>>>>>>>>>>>>>>>>

>>>>>>>>>>>>>>>>>

please give the answer including how u got it ( will add more points if the answer and explanation were clear ) Thanks

y = e2x y = e4x x = 1 x y y = e5x + 3, y = 0, x = 0, x = 0.3 x ex = 2 - x x1 = 0 x2 x3 y = e-x xey + yex = 1 (0, 1) y = f(x) = e2x/8x + 6 y f(x) = e9x/x - 8 , x > 8 x =

Explanation / Answer

Find an equation of the tangent line to the curve xe^(y) + ye^(x) = 1 at the point (0, 1) ?

xe^(y) + ye^(x) = 1

differentiate

e^(y) + xe^(y) *dy/dx + dy/dx * e^(x) + ye^(x) = 0

dy/dx = - [e^(y) + ye^(x)] / [xe^(y) + e^(x)]

put (0, 1)

slope = - [e^(1) + 1e^(0)] / [0 + e^(0)] = - [e + 1]

eq. of the tangent is

y - 1 = -(e+1) *(x - 0)

y = -(e+1)x + 1

Find an equation for the line tangent to the graph of f(x) = e^(2x) /(8x+6) at (1, f(1))

f(1) = e^(2) /14

f(x) = e^(2x) /(8x+6)

differentiate

f'(x) = [ 2e^(2x)*(8x+6) - 8e^(2x) ] /(8x+6)^2

      = [ 2e^(2x)*(8x+2) ] /(8x+6)^2

put (1, f(1)) = (1, e^(2) /14)

slope = [ 2e^(2)*(8+2) ] /(8+6)^2 = 5e^(2) /49

eq. of the tangent is

y - e^(2) /14 = 5e^(2) /49 * (x - 1)

==>

y   = 5e^(2) /49 * (x - 1) + e^(2) /14

The function f(x) = e^(9x) / (x-8)   has one critical number. It is x = ?

f(x) = e^(9x) / (x-8)

diferntiate

f'(x) = [ 9e^(9x) *(x-8) - e^(9x) ] / (x-8)^2

set f'(x) = 0

[ 9e^(9x) *(x-8) - e^(9x) ] / (x-8)^2 = 0

==>

9 *(x-8) - 1 = 0

==>

9x - 73 = 0

x = 73/9

Related Questions

Navigate

• Browse (All)
• Browse #
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.