Determine the intervals on which the function is concave up and concave down
f(x) = 17x^2 = x^4
by differentiating I get
f'(x) = 34x+4x^3
and then I get lost. Can someone finish this up?
f'' > 0 => concave up.
f''(x) = 34+12x^2
since f'' is always positive, f is always concave up.
you can see that because x^2 and x^4 are always positive, and always increasing when x is nonzero.
also, f' = 2x(2x^2+17) which is zero only at x=0. So, only one turning point at (0,0)
Certainly! To determine the intervals on which a function is concave up or concave down, we need to analyze the second derivative of the function. Let's continue with your example.
To find the concavity of the function, we need to differentiate the first derivative, f'(x), with respect to x. Let's calculate the second derivative:
f''(x) = d/dx(34x + 4x^3)
To differentiate 34x, we get 34.
To differentiate 4x^3, we apply the power rule for differentiation. The power rule states that for any term of the form ax^n, the derivative is given by nax^(n-1). Applying this rule, we obtain:
f''(x) = 34 + 12x^2
Now we have the second derivative, f''(x) = 34 + 12x^2. To determine the intervals of concavity, we need to find where the second derivative is positive (concave up) or negative (concave down).
Setting f''(x) = 0, we get:
34 + 12x^2 = 0
Subtracting 34 from both sides, we have:
12x^2 = -34
Dividing both sides by 12, we get:
x^2 = -34/12
Simplifying, we have:
x^2 = -17/6
Since the square of a real number cannot be negative, there are no real solutions to this equation. Therefore, there are no points at which the second derivative changes sign, meaning f''(x) does not cross the x-axis.
Based on this information, the function f(x) = 17x^2 + x^4 is concave up for all real values of x.
To summarize, the function is concave up everywhere.