109. Suppose the derivative of the function y = f(x) is
...

Question

109. Suppose the derivative of the function y = f(x) is

y'=(x-1)^2(x-2).

At what points, if any, does the graph of f have a local minimum, local maximum, or

point of inflection? (Hint: Draw the sign pattern for y'.)

Accepted Answer

Given the function H of Z with its derivative H Z equals Z minus 22 multiplied by 5 minus Z, identify the points where H of Z has a local maximum, local minimum, or point of inflection. Now, to solve this, let's find our critical points first. For our critical point We will set H Z equals a 0. We have Z minus 2 squared multiplied by 5 minus Z equals 0. We can separate the factors to get Z minus 22 equals 0 and 5 minus C equals 0. This gives us two values if we solve. Z equals 2, and Z equals 5. Now, let's look at the sign of our derivative. Now, because our Z equals 2 factor has a multiplicity of 2, this value will always be positive because of the squared. This means we all need to check the sign. Around a point Z equals 5. So, If Z has less than 5 H Z is positive. 40 greater than 5. H Z is negative. We noticed that the derivative is changing between these two values. This means we have a local. Maximum At C equals 5. In particular, this is changing from being positive to negative, which makes it a maximum. We also have no. Local minimum. Now, let's check our inflection points. To find our reflection points, we will set our second derivative equals a zero. So 2nd derivative. In our case, we will use H and Z. And our function right now is. Equals Z minus 2 squared multiplied by 5 minus Z. This is a product rule function. Or our second derivative will be equals to 2 Z minus 2. Multiplied by 5 minus Z. Plus Z minus 2 squad multiplied. By -1. This simplifies To be Z minus 2. Multiplied by 12 minus 3 Z. If we set this equals 0, We get Z minus 2. Equals 0. And 12 minus 3 Z equals 0. This seems we get C equals 2. And Z equals 4. Now we can check the sign. Of our second derivative. We will check it up at the top. Check sign of the 2nd derivative. We have 3 intervals, being negative infinity to 2. To the fore And for to infinity. Let's plug in some arbitrary values. If we plug in 0, for instance, into our second derivative. We get the value -24, which is negative. In our next interval, if we plug in, say, 3, We get the value 3, which is positive. And finally, if you plug in the value on our final interval. Say 5. We get 9, which is negative. We have two sign changes, which means we have Inflection points At Z Equals 2 and 4. We now have our final answer. Local max as Z equals 5, no local min, and inflection points as Z equals 2 and 4. OK, I hope to help me solve the problem. Thank you for watching. Goodbye.

109. Suppose the derivative of the function y = f(x) is
y'=(x-1)^2(x-2).
At what points, if any, does the graph of f have a local minimum, local maximum, or
point of inflection? (Hint: Draw the sign pattern for y'.)

Key Concepts

Derivative and Critical Points

Sign Chart Analysis

Local Extrema and Points of Inflection

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