find the relative maxima

Question

#33. find the relative maxima

y=5ln x-3x

 

#35

find the equation of the tangent line to f(x)=3x+e^x at the point (0,1)

 

#15. find the antiderivative of ∫9e^x/e^x+4

 

#40. find first derivative, and the second derivative , x-value for which relative minimum and maximum, relative-maximum and relative-minimum.

ƒ (x)=x^3/3+3x^2+8x

 3

Answer 1

33. y' = 5/x - 3

 

Set y' = 0 and solve for x.

 

0 = 5/x - 3

 

3x = 5

 

x = 5/3

 

y will have a maximum at x = 5/3.

 

 

The concavity of f(x) is negative for x > 0, so x = 5/3 is a maximum.

 

 

35.  f' = 3 + e^x.  At x = 0, the slope is f'(0) = 3 + 1 = 4.

 

 

The equation of the line through (0,1) is y - 1 = 4(x - 0) or y = 4x + 1.

 

 

 

15.  ∫9e^x/(e^x + 4) dx = 9∫e^x/(e^x + 4) dx

 

 

Let u = e^x + 4, then du = e^x dx.  Substituting,

 

 

9∫du/u = 9ln|u| + C = 9ln|e^x + 4| + C

 

 

 

40.  f(x) = (1/3)x³ + 3x² + 8x + 3

 

First derivative: f' = x² + 6x + 8

 

 

Set f' = 0 to find the critical points (minimum, maximum, inflection points):

 

x² + 6x + 8 = (x + 2)(x + 4) = 0

 

The critical points are x = -2 and x = -4.

 

 

Second derivative: f" = 2x + 6

 

f"(x) gives information on whether a point is a relative minimum (f" > 0), relative maximum (f" < 0), or inflection point (f" = 0). 

 

 

For x = -2, f" = 2(-2) + 6 = 2 > 0.  So, x = -2 would be a relative minimum.

 

For x = -4, f" = 2(-4) + 6 = -2 < 0.  So, x = -4 would be a relative maximum.