Find the extreme values of ƒ(x) = x³ - 3x², and find the area ...

Question

Find the extreme values of ƒ(x) = x³ - 3x², and find the area of the region enclosed by the graph of ƒ and the x-axis.

Accepted Answer

Welcome back everyone. Find the total area of the region bounded by the graph of fx equals x cubed minus 2 x 2 and d x axis, 8/3 square units, b 3/4 square units, C 16/3 square units, and D 4/3 square units. For this problem, let's remember how we find the area bounded by f of x and the x axis. We can show that area is equal to the integral from x1 up to x2 of the absolute value of f of x. D X So what we want to do is simplifying the intersection points. Of f of x and the x axis, right, basically we want to find the x intercepts and do that we said f of x is equal to 0, which means that x cubed minus 2 x 2 is equal to 0. We can factor out x 2 and in parentheses we will have x minus 2. is equal to 0. So from here we can show that one of our solutions is x1 is equal to 0. And the other solution x2 is equal to 2. These are the limits of integration. What we now want to do is check whether our function is above the x axis or below the x axis between x1 and x2, and to do that we can take. The value in between, let's say we take x equals 1. Let's evaluate f at 1, which is going to be 1 cubed minus 2 multiplied by 12, and this is 1 minus 2, which is -1. Since this is less than 0. We know that our function is below the x axis, right? So if we want to drop the absolute value. We're going to take negative of fx. In other words, we are going to set our integral a equals. Integral from 0 to 2 of negative of F of X, if we want to drop the absolute value, right? Based on the definition of piecewise functions. So instead of writing the absolute value of x minus 2x2, which we can also add, that'd be the integral of the absolute value of x cubed minus 2x2. D X Knowing that F of x is negative, we can just drop the absolute value and take the integral from 0 to 2. Of Negative of x cubed minus 2 x 2 dx. We can now distribute the negative sign and integrate from 0 to 2. 2 x 2. Minus X cubed DX. Well done, so now we can apply the power rule, that'd be 2 x cubed divided by 3. Minus x to the power of 4 divided by 4, this is the value of the integral. And we want to evaluate the result from 0 to 2. Let's substitute 2, which gives us 2/3 multiplied by. 2 cube. -2 to the power of 4 divided by 4. And we're going to subtract. 2/3 multiplied by 0 cubed minus. 0 to the power of 4 divided by 4. We can cross out the zero terms and simplify. So we get 2 multiplied by 2 cubed which is 2 to the power of 4, and that's 16. 16 divided by 3 minus 16 divided by 4, which is 4. And now using the common denominator that's. 16 minus 12 divided by 3. Which is 4/3 square units, and this corresponds to the answer of choice d. Thank you for watching.

Find the extreme values of ƒ(x) = x³ - 3x², and find the area of the region enclosed by the graph of ƒ and the x-axis.

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