Mean Value Theorem.

Question

Let f(x) = 4x² -3x -2. What is the value of c in the interval [2,4] guaranteed by the Mean Value Theorem?

Answer 1

We know by the Mean Value Theorem that for a continuous and differentiable function in the interval [a,b], that

f'(c) = [f(b) - f(a)] / (b-a)

Our interval is [2,4], so we will recognize a and b as the following:

a = 2

b = 4

Let's now use the Mean Value Theorem to find the value of the first derivative of our function when the function is located at a point c. We can substitute 4 and 2 into our equation and simplify:

f(4) = 4*(4²) - 3*4 - 2

f(4) = 50

f(2) = 4*(2²) - 3*2 - 2

f(2) = 8

Now we'll take these values for f(4) and f(2) and plug them into the main equation:

f'(c) = [f(4) - f(2)] / (4-2)

f'(c) = [50-8] / 2

f'(c) = 21

So now we know that for the first derivative of our function, there will be some guaranteed point c where f'(c) will be equal to 21. Now we need to find the first derivative of our function to help us solve for c:

f'(x) = 4*2*x - 3 - 0

f'(x) = 8x - 3

Now, simply replace everywhere with an x in our first derivative with a c:

f'(c) = 8c - 3

Finally, equate f'(c) to the value we found from the Mean Value Theorem, and solve for c:

21 = 8c - 3

24 = 8c

We end up with an answer of c = 3

Hope this was helpful!

Answer 2

f'(x) = 8x - 3; f'(c) = 8c - 3; By Mean Value Theorem f'(c) = (f(4) - f(2))/(4 - 2) = (50 - 8)/2 = 21;

we have equation for c: 8c - 3 = 21; 8c = 24, c = 3