Question

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

Answer 1

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

Complete step-by-step solution -
We have a function f with the conditions \[f\left( x+y \right)=f\left( x \right)+f\left( y \right)\forall x,y \And f'\left( 1 \right)=3\].
We want to test the differentiability of f.
We have the condition \[f\left( x+y \right)=f\left( x \right)+f\left( y \right)\forall x,y\]. 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 \[f\left( x \right)=ax\].
If we check the linearity of this function, we observe \[f\left( x+y \right)=a\left( x+y \right)=ax+ay=f\left( x \right)+f\left( y \right)\forall x,y\].
Hence, this satisfies our given condition in the question.
Now, we have \[f'\left( 1 \right)=3\].
We have \[f\left( x \right)=ax\]. We want to evaluate \[\dfrac{d}{dx}f\left( x \right)=\dfrac{d}{dx}\left( ax \right)\]
We know that the differentiation of any function of the form \[y=a{{x}^{n}}\] is such that \[\dfrac{dy}{dx}=an{{x}^{n-1}}\].
Substituting \[n=1\] in the above equation, we get \[\dfrac{d}{dx}f\left( x \right)=\dfrac{d}{dx}\left( ax \right)=a\].
We know that \[f'\left( 1 \right)=3\].
Evaluating \[f'\left( x \right)=a\] at the point \[x=1\], we get \[f'\left( x \right)=a=3\].
Hence, we have \[a=3\].
Thus, the function f is of the form \[f\left( x \right)=3x\] and it satisfies all the given conditions.
Now, we will check the differentiability of f.
We check the differentiability of f at any point \[x=b\] by evaluating that the limit \[\underset{h\to 0}{\mathop{\lim }}\,\dfrac{f\left( b+h \right)-f\left( b \right)}{h}\] exists for all values of b.
Hence, we have \[\underset{h\to 0}{\mathop{\lim }}\,\dfrac{f\left( b+h \right)-f\left( b \right)}{h}=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{3\left( b+h \right)-3b}{h}\].
Solving the above equation, we get \[\underset{h\to 0}{\mathop{\lim }}\,\dfrac{f\left( b+h \right)-f\left( b \right)}{h}=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{3\left( b+h \right)-3b}{h}=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{3h}{h}=\underset{h\to 0}{\mathop{\lim }}\,3=3\].
Now, we evaluate the value of \[f'\left( b \right)\].
We have \[\dfrac{d}{dx}f\left( x \right)=\dfrac{d}{dx}\left( 3x \right)=3\].
Thus, we get \[f'\left( b \right)=3\].
We observe that \[f'\left( b \right)=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{f\left( b+h \right)-f\left( b \right)}{h}=3\].
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.