for f(x)=x^2+kx+k, determine all values of k such that f(x)>0
The minimum of f(x) occurs when
f'(x) = 2x + k = 0, x = -k/2
The value of f(x) at that minimum is
f(-k/2) = k^2/4 -k^2/2 +k = k^2/4 + k
That minimum is >0 if k>0 or k< -4
To determine all values of k such that f(x) > 0, we need to analyze the behavior of the quadratic function f(x) = x^2 + kx + k.
Since the coefficient of x^2 is positive, we know that the graph of f(x) will be a U-shaped parabola that opens upwards.
To find the x-coordinate of the minimum point of the parabola, we can consider the derivative of f(x):
f'(x) = 2x + k
To find the critical points, we set f'(x) equal to zero and solve for x:
2x + k = 0
x = -k/2
The x-coordinate of the minimum point is x = -k/2.
To determine whether f(x) is positive or negative at the minimum point, we substitute x = -k/2 back into f(x):
f(-k/2) = (-k/2)^2 + k(-k/2) + k
= k^2/4 - k^2/2 + k
= k^2/4 + k
To ensure that f(x) is greater than zero (f(x) > 0), we need k^2/4 + k > 0.
Let's solve this inequality:
k^2/4 + k > 0
Firstly, we can see that k can't be equal to zero since k^2/4 would be zero.
Next, we can multiply both sides of the inequality by 4 to eliminate the fraction:
4(k^2/4 + k) > 0 * 4
k^2 + 4k > 0
Now, we can factor the quadratic equation:
k(k + 4) > 0
To determine the sign of k, we can consider the sign of k(k + 4) when k > 0 and when k < -4 (two cases).
Case 1: k > 0
For k > 0, both k and (k + 4) are positive. So, the product k(k + 4) will also be positive.
Case 2: k < -4
For k < -4, k is negative and (k + 4) is also negative. So, the product k(k + 4) will be positive again.
Therefore, the inequality k(k + 4) > 0 holds for all values of k such that k > 0 or k < -4.
In conclusion, all values of k where f(x) > 0 are k > 0 or k < -4.