Need HELP !!! log (1 Viewer)

[sando]

Consider the function y = e^-2x(1 + 2x)

a) Show that the curve has a maximum turning point at (0,1)

b) show that there is a point of inflexion at x = 1/2

 

[withoutaface]

a) Use the product rule to differentiate it, getting y' = 2e^-2x-2e^-2x(1+2x) = -4xe^-2x
Apply it again to find y'' = -4e^-2x + 8xe^-2x

Now y' = -4xe^-2x = -4(0)e^-2(0) = 0 at x=0, hence stationary point.
y = e^-2(0)(1+2(0)) = 1 at x=0, hence the point is (0,1)

y'' = -4e^-2(0) + 8(0)e^-2(0) = -4<0 hence it's a maximum.

b) y'' = -4e^-2(1/2)+8(1/2)e^-2(1/2) = -4/e + 4/e = 0 at x=1/2, hence a point of inflection exists there.

 

[sando]

thanks a load

 

[sando]

some more questions:


1) Consider log_bm² + log_bm³ - log_bm⁴ = 1...... Find a relationship betweenm and b

2) The gradient function at any point (x, f(x)) on a curve is given by f'(x) = e^x + e^-x. Find in terms of x if the curve has a y-intercept of 3....

Sorry bout all the questions guys.... but i just can't think atm

 

[SoulSearcher]

1) log_bm² + log_bm³ - log_bm⁴ = 1
log_b(m²*m³) - log_bm⁴ = 1
log_bm⁵ - log_bm⁴ = 1
log_b(m⁵/m⁴) = 1
log_bm = 1
therefore b = m

 

[sando]

2) anyone ???

 

[word.]

f'(x) = e^x + e^-x -> f(x) = e^x - e^-x + C

if the function has a y-intercept at 3,
then f(x) = 3 when x = 0,

3 = e^0 - e^0 + C -> C = 3,
f(x) = e^x - e^-x + 3

 

[sando]

thanks dude... much appreciated