Question Video: Find All Points in an Interval Which Satisy the Mean Value Theorem for a Quadratic

For the function 𝑓(π‘₯) = π‘₯Β² + 2π‘₯, find all the values of 𝑐 that satisfy the mean value theorem over the interval [βˆ’4, 4].

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Video Transcript

For the function 𝑓 of π‘₯ is equal to π‘₯ squared plus two π‘₯, find all the values of 𝑐 that satisfy the mean value theorem over the closed interval from negative four to four.

We recall that the mean value theorem states that if a function 𝑓 of π‘₯ is continuous on the closed interval from π‘Ž to 𝑏 and is differentiable on the open interval from π‘Ž to 𝑏, then there must exist a point 𝑐 in the open interval from π‘Ž to 𝑏 such that 𝑓 prime evaluated at 𝑐 is equal to 𝑓 evaluated at 𝑏 minus 𝑓 evaluated at π‘Ž all divided by 𝑏 minus π‘Ž. From the question, we can see that we want our function 𝑓 of π‘₯ to be equal to π‘₯ squared plus two π‘₯, and we want to work over the closed interval from negative four to four.

Substituting the information given to us in the question into the mean value theorem gives us that if the function π‘₯ squared plus two π‘₯ is continuous on the closed interval from negative four to four and is differentiable on the open interval from negative four to four. Then we can conclude that there must exist a 𝑐 in the open interval from negative four to four such that 𝑓 prime evaluated at 𝑐 is equal to 𝑓 of four minus 𝑓 of negative four all divided by four minus negative four.

And since the question is asking us to find all of the values of 𝑐 which satisfy the mean value theorem over the closed interval from negative four to four. We then just need to solve this equation given to us for 𝑐 and restrict our answer to the open interval from negative four to four. Because the mean value theorem tells us that this value 𝑐 must exist in this open interval. And the question is asking us to find all values of 𝑐 which satisfy the mean value theorem.

Now, let’s start evaluating the right-hand side of our equation, which is 𝑓 of four minus 𝑓 of negative four divided by four minus negative four. In our numerator, we get 𝑓 evaluated to four, which is equal to four squared plus two multiplied by four. And then, we subtract 𝑓 evaluated at negative four, which is negative four squared plus two multiplied by negative four. And our denominator is four minus negative four, which is equal to eight.

We can then calculate this expression to give us 24 minus eight all divided by eight, which is just equal to two. We can now put this back into our mean value theorem, giving us that we are looking for the values of 𝑐 such that 𝑓 prime of 𝑐 is equal to two. Before we continue any further, let’s check our prerequisites for the mean value theorem. We need that our function is continuous on the closed interval from negative four to four, and it needs to be differentiable on the open interval from negative four to four.

Luckily, our function 𝑓 of π‘₯ is a polynomial. And we know that all polynomials are both continuous and differentiable over all of the real numbers. Now, we want to find our derivative function 𝑓 prime of π‘₯. And we can do this by using the power rule for derivatives, which says that for any constant π‘˜ and constant 𝑛, the derivative of π‘˜ multiplied by π‘₯ to the 𝑛th power is equal to π‘˜ multiplied by 𝑛 multiplied by π‘₯ to the 𝑛 minus one.

Differentiating the first term of π‘₯ squared gives us two π‘₯. And differentiating the second term of two π‘₯ just gives us two since we can just write two π‘₯ as two multiplied by π‘₯ to the first power. Now, we have shown that we can rewrite 𝑓 prime evaluated at 𝑐 as two 𝑐 plus two. And we’ve rewritten 𝑓 of four minus 𝑓 of negative four divided by four minus negative four as just equal to two. This means to find all of the values of 𝑐 that satisfy the mean value theorem over the closed interval from negative four to four, we just need to find the values of 𝑐 in the open interval from negative four to four which satisfy the equation two 𝑐 plus two is equal to two.

And we can calculate that the only solution to the equation two 𝑐 plus two is equal to two, where 𝑐 is in the open interval from negative four to four, is when 𝑐 is equal to zero. Therefore, we have shown that for the function 𝑓 of π‘₯ is equal to π‘₯ squared plus two π‘₯, the only value of 𝑐 which satisfies the mean value theorem over the closed interval from negative four to four is when 𝑐 is equal to zero.

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