f(x) and g(x) are two differentiable functions on [0, 2] such that
f''(x) - g''(x) = 0, f'(1) = 2, g'(1) = 4, f(2) = 3, g(2) = 9
then f(x) - g(x) at x = \[{3 \over 2}\] is
Solutions · 1
Official solution
To find the value of f(x)−g(x) at \[x = \frac{3}{2}\], we need to use the given conditions and properties of differentiable functions.
First, we are told that:
f′′(x)−g′′(x)=0
This implies that:
f′′(x)=g′′(x)
Since the second derivatives of both functions are equal, their difference, f′(x)−g′(x), must be a linear function. Let's denote it as:
f′(x)−g′(x)=k
We'll find the constant k using the initial conditions of the derivatives:
f′(1)=2g′(1)=4
Thus,
f′(1)−g′(1)=2−4=−2
Therefore,
f′(x)−g′(x)=−2
Integrating the above result, we get:
f(x)−g(x)=−2x+C
To determine the constant C, we use the values of the functions at x=2:
f(2)=3g(2)=9
Thus,
f(2)−g(2)=3−9=−6
Therefore,
\[-2 \cdot 2 + C = -6\]
−4+C=−6C=−2
So the expression for f(x)−g(x) is:
f(x)−g(x)=−2x−2
We need to find \[f\left( \frac{3}{2} \right) - g\left( \frac{3}{2} \right)\]:
\[f\left( \frac{3}{2} \right) - g\left( \frac{3}{2} \right) = -2 \cdot \frac{3}{2} - 2\]
=−3−2=−5
Thus, the value of f(x)−g(x) at \[x = \frac{3}{2}\] is -5.