# If for the function

Question:

If for the function $\Phi(x)=\lambda x^{2}+7 x-4, \Phi^{\prime}(5)=97$, find $\lambda$.

Solution:

Given: $\phi(x)=\lambda x^{2}+7 x-4$

Clearly, being a polynomial function, is differentiable everywhere. Therefore the derivative of $\phi$ at $x$ is given by:

$\phi^{\prime}(x)=\lim _{h \rightarrow 0} \frac{\phi(x+h)-\phi(x)}{h}$

$\Rightarrow \phi^{\prime}(x)=\lim _{h \rightarrow 0} \frac{\lambda(x+h)^{2}+7(x+h)-4-\lambda x^{2}-7 x+4}{h}$

$\Rightarrow \phi^{\prime}(x)=\lim _{h \rightarrow 0} \frac{\lambda x^{2}+\lambda h^{2}+2 \lambda x h+7 x+7 h-4-\lambda x^{2}-7 x+4}{h}$

$\Rightarrow \phi^{\prime}(x)=\lim _{h \rightarrow 0} \frac{\lambda h^{2}+2 \lambda x h+7 h}{h}$

$\Rightarrow \phi^{\prime}(x)=\lim _{h \rightarrow 0} \frac{h(\lambda h+2 \lambda x+7)}{h}$

$\Rightarrow \phi^{\prime}(x)=2 \lambda x+7$

It is given $\phi^{\prime}(5)=97$

Thus,

$\phi^{\prime}(5)=10 \lambda+7=97$

$\Rightarrow 10 \lambda+7=97$

$\Rightarrow 10 \lambda=90$

$\Rightarrow \lambda=9$