Question

If \[8f(x) + 6f\left( {\dfrac{1}{x}} \right) = x + 5\]and \[y = {x^2}f(x)\], then what is \[\dfrac{{dy}}{{dx}}\]at \[x = - 1\]?
A. \[0\]
B. \[\dfrac{1}{{14}}\]
C. \[ - \dfrac{1}{{14}}\]
D. \[1\]

Answer 1

Hint: Firstly, we will find the value of \[f(x)\].For that we will first, replace \[x\] by \[\dfrac{1}{x}\] in the given equation \[8f(x) + 6f\left( {\dfrac{1}{x}} \right) = x + 5\] and then we will obtain two equations in terms of \[f(x)\] and \[f\left( {\dfrac{1}{x}} \right)\]. We will then solve the two equations for \[f(x)\]. After solving for \[f(x)\], we will put that value of \[f(x)\] in the given condition \[y = {x^2}f(x)\] to obtain the value of \[y\]. Now, we will get the value of \[y\] in terms of \[x\]i.e. we will get \[y\] as a function of \[x\]. After obtaining the value of \[y\] in terms of \[x\], we will differentiate the value of \[y\]we will obtain in terms of \[x\]with respect to \[x\]. Now, as we have to find the value at \[x = - 1\], we will substitute \[x = - 1\]in the value of \[\dfrac{{dy}}{{dx}}\] obtained.

Complete step by step answer:
We are given,
\[8f(x) + 6f\left( {\dfrac{1}{x}} \right) = x + 5 - - - - - - (1)\]
Now, Replacing \[x\] by \[\dfrac{1}{x}\] in (1), we get
\[ \Rightarrow 8f\left( {\dfrac{1}{x}} \right) + 6f(x) = \dfrac{1}{x} + 5 - - - - - (2)\]
Multiplying first equation by 8,
(1) \[ \times 8 \Rightarrow 8 \times \left[ {\left( {8f(x) + 6f\left( {\dfrac{1}{x}} \right)} \right) = x + 5} \right]\]
\[ = 64f(x) + 48f\left( {\dfrac{1}{x}} \right) = 8x + 40 - - - - - (3)\]

Multiplying second equation by 6,
(2) \[ \times 6 \Rightarrow 6 \times \left[ {8f\left( {\dfrac{1}{x}} \right) + 6f(x) = \dfrac{1}{x} + 5} \right]\]
\[48f\left( {\dfrac{1}{x}} \right) + 36f(x) = \dfrac{6}{x} + 30 - - - - - (4)\]
Now, subtracting (4) from (3), we get
\[ \Rightarrow 64f(x) + 48f\left( {\dfrac{1}{x}} \right) - \left[ {48f\left( {\dfrac{1}{x}} \right) + 36f(x)} \right] = 8x + 40 - \left( {\dfrac{6}{x} + 30} \right)\]
Opening the brackets, we get
\[ \Rightarrow 64f(x) + 48f\left( {\dfrac{1}{x}} \right) - 48f\left( {\dfrac{1}{x}} \right) - 36f(x) = 8x + 40 - \dfrac{6}{x} - 30\]

Cancelling like terms with opposite signs, we get,
\[ \Rightarrow 28f(x) = 8x - \dfrac{6}{x} + 10\]
So, we get the function as,
\[f(x) = \dfrac{1}{{28}}\left( {8x - \dfrac{6}{x} + 10} \right)\]
Now, we have, \[f(x) = \dfrac{1}{{28}}\left( {8x - \dfrac{6}{x} + 10} \right)\]
Substituting the value of \[f(x)\]in \[y = {x^2}f(x)\], we get
\[ \Rightarrow y = {x^2}\left[ {\dfrac{1}{{28}}\left( {8x - \dfrac{6}{x} + 10} \right)} \right]\]
\[ \Rightarrow y = \dfrac{{{x^2}}}{{28}}\left[ {8x - \dfrac{6}{x} + 10} \right]\]
Multiplying \[{x^2}\]inside the bracket
\[ \Rightarrow y = \dfrac{1}{{28}}\left[ {{x^2}\left( {8x - \dfrac{6}{x} + 10} \right)} \right]\]
\[ \Rightarrow y = \dfrac{1}{{28}}\left( {8{x^3} - \dfrac{{6{x^2}}}{x} + 10{x^2}} \right)\]

Simplifying the expression, we get,
\[y = \dfrac{1}{{28}}\left( {8{x^3} - 6x + 10{x^2}} \right)\]
Now, differentiating \[y = \dfrac{1}{{28}}\left( {8{x^3} - 6x + 10{x^2}} \right)\] with respect to \[x\], we have
\[\dfrac{{dy}}{{dx}} = \dfrac{d}{{dx}}\left( {\dfrac{1}{{28}}\left( {8{x^3} - 6x + 10{x^2}} \right)} \right)\]
Taking the constant out of differentiation,
\[ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{1}{{28}}\left[ {\dfrac{d}{{dx}}\left( {8{x^3} - 6x + 10{x^2}} \right)} \right]\]
\[ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{1}{{28}}\left[ {\dfrac{d}{{dx}}8{x^3} - \dfrac{d}{{dx}}6x + \dfrac{d}{{dx}}10{x^2}} \right]\]

Using power rule of differentiation \[\dfrac{d}{{dx}}{x^n} = n \times {x^{n - 1}}\], we get,
\[ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{1}{{28}}\left[ {8 \times 3 \times {x^{3 - 1}} - 6 \times 1 \times {x^{1 - 1}} + 10 \times 2 \times {x^{2 - 1}}} \right]\]
As we know, \[{x^0} = 1\]. So,
\[\dfrac{{dy}}{{dx}} = \dfrac{1}{{28}}\left[ {24{x^2} - 6 + 20x} \right]\]
Hence, we have \[\dfrac{{dy}}{{dx}} = \dfrac{1}{{28}}\left[ {24{x^2} - 6 + 20x} \right]\]
We have to find the value of \[\dfrac{{dy}}{{dx}}\]at \[x = - 1\].
So, we will substitute \[x = - 1\]in \[\dfrac{{dy}}{{dx}} = \dfrac{1}{{28}}\left[ {24{x^2} - 6 + 20x} \right]\]

Therefore, \[\dfrac{{dy}}{{dx}}\]at \[x = - 1\]is
\[\dfrac{{dy}}{{dx}}{|_{x = - 1}} = \dfrac{1}{{28}}\left[ {24{{( - 1)}^2} - 6 + 20( - 1)} \right]\]
We know, \[{(a)^2} = {( - a)^2} = {a^2}\]
\[\dfrac{{dy}}{{dx}}{|_{x = - 1}} = \dfrac{1}{{28}}\left[ {24(1) - 6 - 20)} \right]\]
\[\Rightarrow \dfrac{{dy}}{{dx}}{|_{x = - 1}} = \dfrac{1}{{28}}[24 - 6 - 20]\]
Solving the right side
\[\dfrac{{dy}}{{dx}}{|_{x = - 1}} = \dfrac{1}{{28}}[24 - 26]\]
\[\Rightarrow \dfrac{{dy}}{{dx}}{|_{x = - 1}} = \dfrac{1}{{28}}[ - 2]\]
Simplifying further, we get
\[\therefore \dfrac{{dy}}{{dx}}{|_{x = - 1}} = - \dfrac{1}{{14}}\]
Hence, we get, \[\dfrac{{dy}}{{dx}}\] at \[x = - 1\]equals \[ - \dfrac{1}{{14}}\].

Hence, the correct option is C.

Note: We need to be very careful while finding \[f(x)\]. We must take care of the fact that we have to replace \[x\] by \[\dfrac{1}{x}\] in the whole equation and not just the left hand side. And, then keep in mind to solve for \[f(x)\] and \[f\left( {\dfrac{1}{x}} \right)\] from the obtained and the given equations. Also, we must remember to find the value of \[\dfrac{{dy}}{{dx}}\]at \[x = - 1\] and not just leaving after finding the value of \[\dfrac{{dy}}{{dx}}\]. We must learn the formulas for differentiation very carefully otherwise we will not be able to solve the question.