2018-12-20-120416.sagews

# Kristen Shine
# Final Exam

f(x)=3*x^3-5*x^2-x+8
fprime(x)=diff(f(x),x)
fprime(x)

9*x^2 - 10*x - 1

plot(f(x),-10,10,ymin=-10,ymax=10)

# To find critical points, we set the derivative equal to 0 and solve.

solve(fprime(x)==0,x)
x1=-1/9*sqrt(34) + 5/9
x2=1/9*sqrt(34) + 5/9
n(x1)
n(x2)

y1=f(x1)
y2=f(x2)

[x == -1/9*sqrt(34) + 5/9, x == 1/9*sqrt(34) + 5/9] -0.0923279883161444 1.20343909942726

# To find point of inflection, find second derivative of f, set it equal to 0, and solve.

fdoublep(x)=diff(fprime(x))
solve(fdoublep==0,x)
x3=(5/9)
y3=n(f(x3))

# (x3,y3) is the functions inflection point.

[x == (5/9)]

„#Where is the function f(x) increasing and decreasing?

# Interval -   (-infinity,x1),(x1,x2),(x2,infinity)

# Plug numbers into derivative function from each interval

fprime(-3)
fprime(1/3)
fprime(3)

# From -infinity to x1, the function f(x) is increasing.
# From x1 to x2, the function f(x) is decreasing.
# From x2 to infinity, the function f(x) is increasing.

# When the value of fprime(x) is positive, this means the f(x) function is increasing. Same with negative and decreasing.

File: 
Docstring :

110 -10/3 50

# To find the y coordinates, take f of x1 and x2.
y1=f(x1)
n(y1)
y2=f(x2)
n(y2)
# (x1,y1) is the functions local max
# (x2,y2) is the functions local min

8.04734456316659 4.78393115699802

# plotting function with points -
plot(f(x),-5,5,ymin=-10,ymax=10)+point((x1,y1),size=30)+point((x2,y2),size=30)+point((x3,y3),size=30)