# f(x) = x^4 + ax^2

What is a if f(x) has a local minimum at x=5.
How do I justify the answer.

The first derivative has to be zero at x = 5. You then check what the second derivative is at x = 5. If it is positive then that's ok.

In this case you find that a = -50 and that the second derivative at x = 5 is positive.

## To determine the value of "a" if the function f(x) = x^4 + ax^2 has a local minimum at x = 5, you need to follow a few steps:

1. Take the first derivative of f(x) with respect to x:
f'(x) = d/dx (x^4 + ax^2)

Using the power rule, the derivative of x^n, where n is a constant, is nx^(n-1). Applying this rule to each term in the expression, we get:
f'(x) = 4x^3 + 2ax

2. Set the first derivative f'(x) equal to zero and solve for x:
4x^3 + 2ax = 0

Now, we need to find the value of x when the derivative is equal to zero. In this case, f'(x) = 0 should be true at x = 5.

Substituting x = 5 into the equation, we get:
4(5)^3 + 2a(5) = 0
500 + 10a = 0

To solve for "a", rearrange the equation:
10a = -500

Divide both sides by 10:
a = -50

So, a = -50 in order for f(x) to have a local minimum at x = 5.

Next, you need to verify if the second derivative is positive at x = 5, which helps confirm that it is indeed a local minimum.

To find the second derivative of f(x), you differentiate the first derivative with respect to x:
f''(x) = d^2/dx^2 (4x^3 + 2ax)

Differentiating each term using the power rule, we get:
f''(x) = 12x^2 + 2a

Now, substitute x = 5 into the second derivative expression:
f''(5) = 12(5)^2 + 2a
f''(5) = 300 + 2a

If f''(5) > 0, then the second derivative is positive, indicating a local minimum at x = 5.

Substituting a = -50, we have:
f''(5) = 300 + 2(-50)
f''(5) = 300 - 100
f''(5) = 200

Since f''(5) = 200 is positive, it confirms that a = -50 results in a local minimum at x = 5.

Therefore, a = -50 and the second derivative f''(5) = 200 justifies this answer.