Let f(x)=e^(x2). By the Mean Value Theorem, there exists a point c such that
f′(c)=f(6)−f(4)6−4
with c between 4 and 6. Approximate c to within two decimal places; we're asking you to approximate c, because you can't "solve" for c.
Your statement of
f′(c)=f(6)−f(4)6−4
desperately needs brackets to say
f′(c)= ( f(6)−f(4) )/(6−4)
f(6) = e^36 = 4.31123x10^15 , rather large
f(4) = e^16 = only 8886110.52
( f(6) - f(4) )/2 = 2.1556x10^15
To approximate the value of c using the given equation and the Mean Value Theorem, we need to find the derivative of the function f(x) = e^(x^2) and then substitute the values f(6) and f(4) into the equation.
Let's start by finding the derivative of f(x):
f(x) = e^(x^2)
Using the chain rule, we have:
f'(x) = 2x * e^(x^2)
Now, let's substitute the values f(6) = f(4) into the equation:
f'(c) = [f(6) - f(4)] / (6 - 4)
Substituting the function values:
f'(c) = [e^(6^2) - e^(4^2)] / (6 - 4)
Now, we need to approximate the value of c by solving the equation. The equation cannot be solved algebraically, so we'll need to use numerical methods to obtain an approximation.
One common numerical method is the bisection method:
1. Start with an interval [a, b] that contains the solution c. We know that c is between 4 and 6, so we can choose [4, 6] as our interval.
2. Calculate the value of f'(c) for the midpoint of the interval, (a + b) / 2.
3. If f'(c) is very close to zero (within a certain tolerance), the midpoint is an approximation for c.
4. If f'(c) is positive, the solution should be in the left half of the interval. Set b = (a + b) / 2.
5. If f'(c) is negative, the solution should be in the right half of the interval. Set a = (a + b) / 2.
6. Repeat steps 2-5 until the value of f'(c) is within the desired tolerance.
By repeatedly applying the bisection method, you can approximate the value of c to within two decimal places.