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CONTINUITY AND DIFFERENTIABILITY — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsCONTINUITY AND DIFFERENTIABILITY5 Marks
Question
If f is defined by $\text{f(x)}=\text{x}^2-4\text{x}+7,$ show that $\text{f}'(5)=2\text{f}'\Big(\frac{7}{2}\Big).$
✓

Answer

$f(x) = x^2 - 4x + 7$ is a polynomial function, So it is differentiable everywhere.
$\text{f}'(5)=\lim_\limits{\text{h}\rightarrow0}\frac{\text{f}(5+\text{h})-\text{f}(5)}{\text{h}}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\big\{(5+\text{h})^2-4(5+\text{h})+7\big\}-[25-20+7]}{\text{h}}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\text{h}^2+25+10\text{h}-20-4\text{h}+7-12}{\text{h}}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\text{h}^2+6\text{h}}{\text{h}}$
$=\lim_\limits{\text{h}\rightarrow0}(\text{h}+6)$
$=6$
$\text{f}'\Big(\frac{7}{2}\Big)=\lim_\limits{\text{h}\rightarrow0}\frac{\text{f}\Big(\frac{7}{2}+\text{h}\Big)^2-\text{f}\Big(\frac{7}{2}\Big)}{\text{h}}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\Big[\Big(\frac{7}{2}+\text{h}\Big)^2-4\Big(\frac{7}{2}+\text{h}\Big)+7\Big]-\Big[\Big(\frac{7}{2}\Big)^2-4\Big(\frac{7}{2}\Big)+7\Big]}{\text{h}}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\Big[\frac{49}{2}+\text{h}^2+7\text{h}-14-4\text{h}+7\Big]-\Big[\frac{49}{2}-14+7\Big]}{\text{h}}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\frac{49}{2}+\text{h}^2+7\text{h}-14-4\text{h}+7-\frac{49}{2}-14+7}{\text{h}}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\text{h}^2+3\text{h}}{\text{h}}$
$=\lim_\limits{\text{h}\rightarrow0}(\text{h}+3)$
$=3$
$\text{f}'(5)=6$
$=2(3)$
$=\text{f}'(5)=2\text{f}'\Big(\frac{7}{2}\Big)$

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