If f ( x ) is a differentiable function and g ( x ) is a double differentiable function such that | f ( x ) | ≤ 1 and f ' ( x ) = g ( x ) . If f 2 ( 0 ) + g 2 ( 0 ) = 9 . Prove that there exists some c ∈ ( – 3 , 3 ) such that g ( c ) . g '' ( c ) < 0.
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In this problem, we are given that f(x) is a differentiable function with absolute value less than or equal to 1, and g(x) is its derivative. We also know that f squared of 0 plus g squared of 0 equals 9. We need to prove that there exists some value c in the interval negative 3 to 3 such that the product of g of c and the second derivative of g at c is negative. We'll approach this by contradiction, assuming no such point exists, and then show this leads to an impossible situation.
Let's start our proof by contradiction. We'll assume that g of x times g double prime of x is greater than or equal to zero for all x in the interval negative 3 to 3. Now, let's define a new function h of x as g squared of x. When we compute the second derivative of h, we get 2 times g prime squared plus 2 times g times g double prime. Since we assumed g times g double prime is non-negative, h double prime is greater than or equal to 2 times g prime squared, which is always non-negative. This means h double prime is non-negative throughout the interval, which implies that h of x, or g squared of x, is a convex function on the interval negative 3 to 3. This convexity property will be crucial for our contradiction.
Now, let's derive some bounds on g of x. From our given conditions, we know that f squared of 0 plus g squared of 0 equals 9, and the absolute value of f of x is at most 1. This means f squared of 0 is at most 1, which implies g squared of 0 is at least 8. Next, using the Mean Value Theorem on the interval from 0 to 3, there exists some point xi_1 between 0 and 3 such that f of 3 minus f of 0 equals g of xi_1 times 3. Since the absolute values of f of 0 and f of 3 are both at most 1, their difference has absolute value at most 2. This gives us that the absolute value of g of xi_1 is at most 2/3, so g squared of xi_1 is at most 4/9. Similarly, applying the Mean Value Theorem on the interval from negative 3 to 0, we find another point xi_2 where g squared of xi_2 is also at most 4/9. These bounds will create a contradiction with the convexity of g squared.
Now we'll use the convexity of g squared to reach a contradiction. For a convex function, the slope of the secant line from point a to point x is less than or equal to the derivative at x. Looking at the secant line from xi_2 to 0, we get that the slope is less than or equal to the derivative of g squared at 0, which is 2 times g of 0 times g prime of 0. Plugging in our bounds, we get that g of 0 times g prime of 0 must be greater than 0. However, when we look at the secant line from 0 to xi_1, the convexity property gives us that the slope is greater than or equal to the derivative at 0. Using our bounds again, we get that g of 0 times g prime of 0 must be less than 0. This is a contradiction! The same quantity cannot be both positive and negative. Therefore, our initial assumption must be false.
Let's summarize our proof. We started by assuming that g of x times g double prime of x is greater than or equal to zero for all x in the interval negative 3 to 3. This led us to conclude that g squared of x is a convex function. Using the given conditions that the absolute value of f of x is at most 1 and f squared of 0 plus g squared of 0 equals 9, we derived that g squared of 0 is at least 8. Then, using the Mean Value Theorem, we found points xi_1 and xi_2 where g squared is at most 4/9. Applying properties of convex functions to the secant lines, we reached a contradiction: g of 0 times g prime of 0 would have to be both positive and negative, which is impossible. Therefore, our initial assumption must be false, and there must exist some point c in the interval negative 3 to 3 where g of c times g double prime of c is negative. This proof demonstrates the power of contradiction in establishing existence results in calculus.