Question
If for function $\phi(\text{x})=\lambda\text{x}^2+7\text{x}-4, \phi(5)=97,$ find $\lambda.$
✓
Answer
Given: $\phi(\text{x})=\lambda\text{x}^2+7\text{x}-4$
Clearly, being a polynomial function, is differentiable everywhere. Therefore the derivative of $\phi$ at x is given by:
$\phi'(\text{x})=\lim_\limits{\text{h}\rightarrow0}\frac{\phi(\text{x}+\text{h})-\phi(\text{x})}{\text{h}}$
$\Rightarrow\phi'(\text{x})=\lim_\limits{\text{h}\rightarrow0}\frac{\lambda(\text{x}+\text{h})^2+7(\text{x}+\text{h})-4-\lambda\text{x}^2-7\text{x}+4}{\text{h}}$
$\Rightarrow\phi'(\text{x})=\lim_\limits{\text{h}\rightarrow0}\frac{\lambda(\text{x}+\text{h})^2+7(\text{x}+\text{h})-4-\lambda\text{x}^2-7\text{x}+4}{\text{h}}$
$\Rightarrow\phi'(\text{x})=\lim_\limits{\text{h}\rightarrow0}\frac{\lambda\text{h}^2+2\lambda\text{xh}+7\text{h}}{\text{h}}$
$\Rightarrow\phi'(\text{x})=\lim_\limits{\text{h}\rightarrow\infty}\frac{\text{h}(\lambda\text{h}+2\lambda\text{x}+7)}{\text{h}}$
$\Rightarrow\phi'(\text{x})=2\lambda\text{x}+7$
It is given $\phi'(5)=97$
Thus, $\phi'(5)=10\lambda+7=97$
$\Rightarrow10\lambda+7=97$
$\Rightarrow10\lambda=90$
$\Rightarrow\lambda=9$
Clearly, being a polynomial function, is differentiable everywhere. Therefore the derivative of $\phi$ at x is given by:
$\phi'(\text{x})=\lim_\limits{\text{h}\rightarrow0}\frac{\phi(\text{x}+\text{h})-\phi(\text{x})}{\text{h}}$
$\Rightarrow\phi'(\text{x})=\lim_\limits{\text{h}\rightarrow0}\frac{\lambda(\text{x}+\text{h})^2+7(\text{x}+\text{h})-4-\lambda\text{x}^2-7\text{x}+4}{\text{h}}$
$\Rightarrow\phi'(\text{x})=\lim_\limits{\text{h}\rightarrow0}\frac{\lambda(\text{x}+\text{h})^2+7(\text{x}+\text{h})-4-\lambda\text{x}^2-7\text{x}+4}{\text{h}}$
$\Rightarrow\phi'(\text{x})=\lim_\limits{\text{h}\rightarrow0}\frac{\lambda\text{h}^2+2\lambda\text{xh}+7\text{h}}{\text{h}}$
$\Rightarrow\phi'(\text{x})=\lim_\limits{\text{h}\rightarrow\infty}\frac{\text{h}(\lambda\text{h}+2\lambda\text{x}+7)}{\text{h}}$
$\Rightarrow\phi'(\text{x})=2\lambda\text{x}+7$
It is given $\phi'(5)=97$
Thus, $\phi'(5)=10\lambda+7=97$
$\Rightarrow10\lambda+7=97$
$\Rightarrow10\lambda=90$
$\Rightarrow\lambda=9$
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