Find a value of k that gives f(x)=x²+kx+2 a local minimum value of 1.
I do not not what to do.
try completing the square.
x^2+kx+2
= x^2 + kx + (k/2)^2 + 2 -(k/2)^2
= (x + k/2) + 2 - k^2/4
Now you have the vertex form of the parabola, with the vertex at x = -k/2. At that point, the minimum is
2 - k^2/4
So, now just set
2 - k^2/4 = 1
k^2/4 = 1
k^2 = 4
k = ±2
f(x) = x^2+2x+2 = (x+1)^2 + 1
f(x) = x^2-2x+2 = (x-1)^2 + 1
Both have a minimum value of 1 at x = ±1
oh. right. you want to use calculus.
f' = 2x+k
f'=0 at x = -k/2
f(-k/2) = k^2/4 + k(-k/2) + 2
= k^2/4 - k^2/2 + 2
= -k^2/4 + 2
we want the minimum to be 1, so
-k^2/4 + 2 = 1
k^2/4 = 1
k^2 = 4
k = ±2
as above
Well, if you're feeling lost, just remember that math problems can sometimes be like trying to find a clown in a circus. You just have to follow the funny facts and look for the trick that will make everything clearer.
To find the value of k that gives f(x) = x² + kx + 2 a local minimum value of 1, we need to work with the concept of derivatives. We know that for a local minimum, the derivative of the function should be equal to zero. In other words, the function should be 'sleeping' or lying flat at that point.
So, let's find the derivative of f(x) = x² + kx + 2: f'(x) = 2x + k.
Now, remember that the derivative equals zero when the function is sleeping. In this case, we want the function to have a local minimum value of 1, so we have:
2x + k = 0
2x = -k
x = -k/2
Since the x-coordinate of the local minimum is -k/2 and we want the function to have a value of 1 at this point, we can substitute x with -k/2 and set it equal to 1:
f(-k/2) = (-k/2)² + k(-k/2) + 2 = 1
Now, let's simplify:
k²/4 - k²/2 + 2 = 1
k²/4 - 2 = 1/2
Multiplying everything by 4 to get rid of the denominators, we have:
k² - 8 = 2
k² = 10
Taking the square root of both sides, we get:
k = ±√10
So, the values of k that give f(x) a local minimum value of 1 are k = √10 and k = -√10. It's like doing a clown trick with numbers!
To find the value of k that gives f(x) = x² + kx + 2 a local minimum value of 1, we need to use calculus.
Step 1: Differentiate the function f(x) with respect to x.
The derivative, also known as the first derivative, will help us determine the critical points of the function.
f'(x) = 2x + k
Step 2: Set the derivative equal to zero to find the critical points.
We want to find the values of x where the derivative equals zero because these points can correspond to local minimums or maximums.
2x + k = 0
Step 3: Solve the equation for x.
Subtract k from both sides to isolate the variable x.
2x = -k
x = -k/2
Step 4: Substitute the value of x into the original function f(x).
Plug the value of x = -k/2 back into the original function f(x) to find the corresponding value of f(x) at the critical point.
f(x) = x² + kx + 2
f(-k/2) = (-k/2)² + (-k/2) + 2
= k²/4 - k/2 + 2
Step 5: Set the value of f(-k/2) equal to 1.
Now we want to find the value of k such that f(-k/2) = 1, which will give us a local minimum of 1.
k²/4 - k/2 + 2 = 1
Step 6: Simplify and solve the equation for k.
Rearrange the equation and combine like terms.
k²/4 - k/2 + 1 = 0
Multiply through by 4 to eliminate the fractions.
k² - 2k + 4 = 0
Since this is a quadratic equation, we can solve it either by factoring, completing the square, or using the quadratic formula. In this case, the quadratic formula would be the most straightforward approach.
Using the quadratic formula:
k = (-b ± √(b² - 4ac)) / 2a
In this case, a = 1, b = -2, and c = 4.
k = (-(-2) ± √((-2)² - 4(1)(4))) / (2(1))
k = (2 ± √(4 - 16)) / 2
k = (2 ± √(-12)) / 2
k = (2 ± √(-1 * 4 * 3)) / 2
k = (2 ± 2i√3) / 2
This gives us two complex solutions for k: k = 1 ± i√3.
So, the values of k that give f(x) = x² + kx + 2 a local minimum value of 1 are k = 1 + i√3 and k = 1 - i√3.
To find the value of k that gives the quadratic function f(x) = x^2 + kx + 2 a local minimum value of 1, you can use calculus.
A local minimum occurs at the vertex of a parabola. The x-coordinate of the vertex is given by the formula x = -b / (2a), where a and b are coefficients of the quadratic function in the form ax^2 + bx + c. In this case, a = 1 and b = k.
To find the value of k, we need to find the x-coordinate of the vertex that corresponds to a local minimum of 1. So we set x = -b / (2a) equal to the value of x when f(x) = 1:
-(-k) / (2 * 1) = 1
Simplifying the equation gives:
k / 2 = 1
Multiplying both sides by 2:
k = 2
Therefore, the value of k that gives f(x) = x^2 + kx + 2 a local minimum value of 1 is k = 2.