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Question: If a function is given as \[f\left( x+y \right)=f\left( x \right)+f\left( y \right)\forall x,y \And ...

If a function is given as f(x+y)=f(x)+f(y)∀x,y&f′(1)=3f\left( x+y \right)=f\left( x \right)+f\left( y \right)\forall x,y \And f'\left( 1 \right)=3, then test the differentiability of f(x)f\left( x \right).

Explanation

Solution

Hint: Use linearity of the functions to find the exact functions and then test the differentiability of f(x)f\left( x \right) by evaluating the value of lim⁡h→0 f(b+h)−f(b)h\underset{h\to 0}{\mathop{\lim }}\,\dfrac{f\left( b+h \right)-f\left( b \right)}{h}.

Complete step-by-step solution -
We have a function f with the conditions f(x+y)=f(x)+f(y)∀x,y&f′(1)=3f\left( x+y \right)=f\left( x \right)+f\left( y \right)\forall x,y \And f'\left( 1 \right)=3.
We want to test the differentiability of f.
We have the condition f(x+y)=f(x)+f(y)∀x,yf\left( x+y \right)=f\left( x \right)+f\left( y \right)\forall x,y. Thus, we can see that f is a linear function.
Hence, we can assume that f is a function of the form of a polynomial with degree 1 such that f(x)=axf\left( x \right)=ax.
If we check the linearity of this function, we observe f(x+y)=a(x+y)=ax+ay=f(x)+f(y)∀x,yf\left( x+y \right)=a\left( x+y \right)=ax+ay=f\left( x \right)+f\left( y \right)\forall x,y.
Hence, this satisfies our given condition in the question.
Now, we have f′(1)=3f'\left( 1 \right)=3.
We have f(x)=axf\left( x \right)=ax. We want to evaluate ddxf(x)=ddx(ax)\dfrac{d}{dx}f\left( x \right)=\dfrac{d}{dx}\left( ax \right)
We know that the differentiation of any function of the form y=axny=a{{x}^{n}} is such that dydx=anxn−1\dfrac{dy}{dx}=an{{x}^{n-1}}.
Substituting n=1n=1 in the above equation, we get ddxf(x)=ddx(ax)=a\dfrac{d}{dx}f\left( x \right)=\dfrac{d}{dx}\left( ax \right)=a.
We know that f′(1)=3f'\left( 1 \right)=3.
Evaluating f′(x)=af'\left( x \right)=a at the point x=1x=1, we get f′(x)=a=3f'\left( x \right)=a=3.
Hence, we have a=3a=3.
Thus, the function f is of the form f(x)=3xf\left( x \right)=3x and it satisfies all the given conditions.
Now, we will check the differentiability of f.
We check the differentiability of f at any point x=bx=b by evaluating that the limit lim⁡h→0 f(b+h)−f(b)h\underset{h\to 0}{\mathop{\lim }}\,\dfrac{f\left( b+h \right)-f\left( b \right)}{h} exists for all values of b.
Hence, we have lim⁡h→0 f(b+h)−f(b)h=lim⁡h→0 3(b+h)−3bh\underset{h\to 0}{\mathop{\lim }}\,\dfrac{f\left( b+h \right)-f\left( b \right)}{h}=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{3\left( b+h \right)-3b}{h}.
Solving the above equation, we get lim⁡h→0 f(b+h)−f(b)h=lim⁡h→0 3(b+h)−3bh=lim⁡h→0 3hh=lim⁡h→0 3=3\underset{h\to 0}{\mathop{\lim }}\,\dfrac{f\left( b+h \right)-f\left( b \right)}{h}=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{3\left( b+h \right)-3b}{h}=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{3h}{h}=\underset{h\to 0}{\mathop{\lim }}\,3=3.
Now, we evaluate the value of f′(b)f'\left( b \right).
We have ddxf(x)=ddx(3x)=3\dfrac{d}{dx}f\left( x \right)=\dfrac{d}{dx}\left( 3x \right)=3.
Thus, we get f′(b)=3f'\left( b \right)=3.
We observe that f′(b)=lim⁡h→0 f(b+h)−f(b)h=3f'\left( b \right)=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{f\left( b+h \right)-f\left( b \right)}{h}=3.
Hence, we observe that the given limit exists for all values of b.
Thus, we see that f is a linear differentiable function.

Note: One must know the exact formula required for the differentiability of f. Also, it’s very necessary to observe that f if a linear function. Otherwise, we won’t be able to solve this question. We can also assume any other linear function which satisfies the given condition and check its differentiability.