How do you find the maximum, minimum and inflection points for each function #f(x)= -14x^3 + 19x^3 - x - 2#?

Question

Ezra Adams · Accepted Answer

Critical points occur at #f'(x_0)=0#, and may be maxima or minima. Inflection points can only occur at #f''(x_0)=0#. See explanation

First off, note that this is a third order function. A function of third order has no global maxima or minima; at one end it trends towards positive infinity, and at the other it trends towards negative infinity,. Therefore, no absolute minima or maxima exist. If this were a function of order 2 Local maxima and minima are types of critical points. Critical points are points where the function is neither increasing nor decreasing, i.e. where the slope of the line tangent to the function at that point is 0. Since the derivative of a function is the slope of its tangent line, that means that the critical points occur at the zeroes of the derivative. This explanation will proceed under the belief that you meant to write #−14x^3+19x^2−x−2#. If the second term is indeed #19x^3# as you wrote in your problem, simply add the first and second terms together and then take the derivative of the function. Recall that the power rule states given a function #f(x) = ax^n + bx^(n-1) + ...+kx+m, (df)/dx = nax^(n-1) + (n-1)bx^(n-2) + ...+k#. Thus here: #d/dx (-14x^3+19x^2-x-2) = -42x^2+38x-1#. Using the quadratic formula and a calculator, we obtain zeroes at #x approx 0.027# and #x approx 0.878# To determine what types of critical points these are, choose an arbitrarily small number #epsilon# and find the derivative at the points #x_0 +- epsilon#, with x0 being the zeroes. #f'(0.027+epsilon) = -42(.027+epsilon)^2 + 38(0.027+epsilon)-1 approx -0.031 -2.268epsilon -42epsilon^2 + 1.026 + 38epsilon-1# Knowing that the non epsilon terms will cancel out.. #approx -2.268 epsilon - 42epsilon^2 + 38epsilon = epsilon (-42epsilon+ 35.732)# This interior term will remain positive as epsilon approaches zero whether or not epsilon is positive or negative, thus only the epsilon on the outside determines our sign. Therefore, for a negative epsilon (i.e. #x-epsilon#), our derivative is negative, and for a positive epsilon it is positive. Thus, the derivative shifts from negative to positive as it crosses this zero, meaning that this is a local minimum . Checking the second zero is left to the student, but you should be able to determine that it is a local maximum, meaning that the derivative changes from positive to negative as it crosses the 0. Inflection points are the points where the second derivative is equal to 0. You can think of them as the critical points of the derivative , but this may be somewhat confusing. Using the power rule; #(d^2f)/(dx^2) = -84x+38# Only one zero exists, at #x=(38/84)#. Thus this is the only possible point of inflection. Since the equation has a negative slope, it is changing from positive to negative at the zero in question, meaning that the function changes from concave up (i.e. convex) to concave down (i.e. concave) at #x=38/84#

Charles Arocha · Answer

undefined To find the maximum and minimum points, we first find the critical points by setting the derivative equal to zero and solving for $x$. Then, we use the second derivative test to determine if each critical point is a maximum or minimum. For inflection points, we find where the second derivative equals zero and test the concavity of the function around those points. Let's start by finding the derivatives of the given function:

$f(x) = -14x^3 + 19x^2 - x - 2$

$f'(x) = -42x^2 + 38x - 1$

$f''(x) = -84x + 38$

Now, set the first derivative equal to zero and solve for $x$ to find critical points:

$-42x^2 + 38x - 1 = 0$

Next, use the quadratic formula to solve for $x$. Then, plug the values of $x$ into the second derivative to classify each critical point as a maximum or minimum. Finally, for inflection points, set the second derivative equal to zero and solve for $x$.