For what values of c does the function f (x) = cx + arctan x have exactly one local maximum and one local minimum?
where does the 4-4c^2 come from
review the discriminant of a quadratic.
Think back to Algebra I
b^2 - 4ac? Where does the 4 come from, as well is the value of a 1?
To find the values of c that make the function f(x) = cx + arctan(x) have exactly one local maximum and one local minimum, we need to analyze the properties of the function.
A function has a local maximum or minimum at a particular point if the derivative of the function changes sign at that point. So, we first need to calculate the derivative of f(x) with respect to x.
Let's start by finding the derivative of the function, f'(x):
f'(x) = c + (1/(1 + x^2))
Now, for the function to have exactly one local maximum and one local minimum, the derivative must change sign twice around this point. This means that f'(x) must be positive, then negative, and then positive again (or vice versa).
Let's analyze the derivative more closely.
1. To determine where f'(x) is positive, we need to solve the inequality f'(x) > 0.
c + (1/(1 + x^2)) > 0
To isolate x, subtract c from both sides and multiply both sides by (1 + x^2):
1/(1 + x^2) > -c
Since (1 + x^2) is always positive, we can multiply both sides by (1 + x^2) without changing the inequality direction:
1 > -c(1 + x^2)
Now, we have two cases to consider:
a) If c < 0, the inequality becomes:
1 > c(1 + x^2)
Since c < 0, the inequality holds true for all x. So, for c < 0, f'(x) > 0 for all values of x, meaning the function f(x) = cx + arctan(x) has no local maximum or minimum.
b) If c > 0, the inequality becomes:
1 < c(1 + x^2)
Divide both sides by (1 + x^2):
1/(1 + x^2) < c
This means that f'(x) > 0 when 1/(1 + x^2) < c.
2. To determine where f'(x) is negative, we need to solve the inequality f'(x) < 0.
c + (1/(1 + x^2)) < 0
Subtract c from both sides:
1/(1 + x^2) < -c
Notice that this inequality is the same as the one we obtained in case b) above. So, for f'(x) to be negative, we need to have the same condition as f'(x) being positive:
1/(1 + x^2) < c
This means that f'(x) < 0 when 1/(1 + x^2) < c.
To summarize, the condition for f(x) = cx + arctan(x) to have exactly one local maximum and one local minimum is:
1/(1 + x^2) < c, where c > 0
Therefore, the values of c that satisfy this condition are c > 0.
f'(x) = c + 1/(x^2+1) = (cx^2+x+c)/(1+x^2)
One max and one min means that cx^2 + x + c has two real roots, so
4-4c^2 > 0
c^2 < 1
Clearly, c>0 has no extrema, since both cx and arctanx are always increasing
So, -1 < c < 0