Use a graph of f(x) = 3 e^{-8 x^2} to estimate the x-values of any critical points and inflection points of f(x). critical points x=
Inflection points x=
Next, use derivatives to find the x-values of any critical points and inflection points exactly.
critical points x=
Inflection points x=
To estimate the x-values of the critical points and inflection points of the function f(x) = 3e^(-8x^2) using a graph, we need to examine the behavior of the graph around these points.
Critical points correspond to locations where the slope of the function is equal to zero or is undefined. For this function, the critical points occur where the derivative is equal to zero or undefined. Since we are given the function, we can find the derivative of f(x) to determine the critical points.
To find the derivative, we differentiate f(x) with respect to x:
f'(x) = d/dx (3e^(-8x^2))
To calculate f'(x), we can follow the rules of differentiation. The derivative of e^u, where u is a function of x, is given by e^u * u'. Applying this rule, we have:
f'(x) = 3 * (-16x) * e^(-8x^2)
= -48x * e^(-8x^2)
Now, let's solve f'(x) = 0 to find the x-values of the critical points.
-48x * e^(-8x^2) = 0
This equation is satisfied when x = 0, as the factor -48x becomes zero. Therefore, x = 0 is a critical point of f(x).
To estimate the inflection points, we need to examine the concavity of the graph. An inflection point occurs where the concavity of the graph changes. This is determined by the second derivative of the function. Let's find the second derivative of f(x).
f''(x) = d^2/dx^2 (-48x * e^(-8x^2))
Differentiating, we find:
f''(x) = 48(64x^2 - 1) * e^(-8x^2)
Now, set f''(x) = 0 to find the x-values of the inflection points:
48(64x^2 - 1) * e^(-8x^2) = 0
This equation is satisfied when x^2 = 1/64. Solving for x, we find two values: x = 1/8 and x = -1/8. These are the approximate x-values of the inflection points.
To find the x-values of the critical points and inflection points exactly using derivatives, we need to solve the equations f'(x) = 0 and f''(x) = 0.
From earlier, we found that f'(x) = -48x * e^(-8x^2) = 0. Setting this equation equal to zero, we have:
-48x * e^(-8x^2) = 0
Since e^(-8x^2) is always positive and nonzero, the only way for the product to be zero is if -48x = 0, which gives x = 0. Hence, the critical point is x = 0.
Similarly, for the equation f''(x) = 0, we had:
48(64x^2 - 1) * e^(-8x^2) = 0
Since e^(-8x^2) is always positive and nonzero, we solve 64x^2 - 1 = 0, which gives x^2 = 1/64. Taking the square root, we find two solutions: x = 1/8 and x = -1/8. These are the exact x-values of the inflection points.
Therefore, the critical point of f(x) is x = 0, and the inflection points are x = 1/8 and x = -1/8.