Question

If for the function 

\[\Phi \left( x \right) = \lambda x^2 + 7x - 4, \Phi'\left( 5 \right) = 97, \text { find } \lambda .\]

Answer 1

Given:  

\[\phi(x) = \lambda x^2 + 7x - 4\]

Clearly, being a polynomial function, is differentiable everywhere. Therefore the derivative of 

\[\phi(x) = \lambda x^2 + 7x - 4\]

\[x\] is given by:

\[\phi'(x) = \lim_{h \to 0} \frac{\phi(x + h) - \phi(x)}{h}\]

\[ \Rightarrow \phi'(x) = \lim_{h \to 0} \frac{\ \lambda (x + h )^2 + 7(x + h) - 4 - \lambda x^2 - 7x + 4}{h}\]

\[ \Rightarrow \phi'(x) = \lim_{h \to 0} \frac{\ \lambda  x^2 + \lambda  h^2 + 2\lambda xh + 7x + 7h - 4 -\lambda x^2 - 7x + 4}{h}\]

\[ \Rightarrow \phi'(x) = \lim_{h \to 0} \frac{\lambda h^2 + 2\lambda xh + 7h}{h}\]

\[ \Rightarrow \phi'(x) = \lim_{h \to 0} \frac{h(\lambda h + 2\lambda x + 7)}{h}\]

\[ \Rightarrow \phi'(x) = 2\lambda x + 7\]

It is given 

\[\phi'(5) = 97\]

Thus,

\[\phi'(5) = 10\lambda + 7 = 97\]

\[ \Rightarrow 10\lambda  + 7 = 97\]

\[ \Rightarrow 10\lambda  = 90\]

\[ \Rightarrow\lambda  = 9\]