Question

1. On what interval is the function f(x)=x^3-4x^2+5x concave upward?

2. For what values of x dies the graph of f(x)=2x^3-3x^2-6x+87 have a horizontal tangent?

Is there no solution because the equation can't be factored?


3. At what point on the curve y=1+2e^x - 3x is the tangent line parallel to the line 3x-y=5?

Answer 1

#1
f' = 3x^2-8x+5
f" = 6x-8
f is concave up when f" > 0

#2
f' = 6x^2-6x-6
maybe it cannot be factored, but since the discriminant is positive, it has real solutions. Use the quadratic formula to find them.

#3 you want the slope to be 3.
y' = 2e^x - 3
so,
2e^x - 3 = 3
2e^x = 6
e^x = 3
x = ln 3
y(ln3) = 1 + 6 - 3ln3 = 7-ln27
So, the tangent line is

y = 3(x-ln3) + 7-ln27

See the graph at

http://www.wolframalpha.com/input/?i=plot+y%3D1%2B2e^x+-+3x%2C+y%3D3%28x-ln3%29+%2B+7-ln27