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Question
If \[f\left( x \right) = x^3 + 7 x^2 + 8x - 9\]
, find f'(4).
Answer in Brief
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Solution
Given:
\[f(x) = x^3 + 7 x^2 + 8x - 9\]
Clearly, being a polynomial function, is differentiable everywhere. Therefore the derivative of
\[f\] at
\[x\] is given by:
\[f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h}\]
\[ \Rightarrow f'(x) = \lim_{h \to 0} \frac{(x + h )^3 + 7(x + h )^2 + 8(x + h) - 9 - x^3 - 7 x^2 - 8x + 9}{h}\]
\[ \Rightarrow f'(x) = \lim_{h \to 0} \frac{x^3 + h^3 + 3 x^2 h + 3x h^2 + 7 x^2 + 7 h^2 + 14xh + 8x + 8h - 9 - x^3 - 7 x^2 - 8x + 9}{h}\]
\[ \Rightarrow f'(x) = \lim_{h \to 0} \frac{h^3 + 3 x^2 h + 3x h^2 + 7 h^2 + 14xh + 8h}{h}\]
\[ \Rightarrow f'(x) = \lim_{h \to 0} \frac{h( h^2 + 3 x^2 + 3xh + 7h + 14x + 8)}{h}\]
\[ \Rightarrow f'(x) = \lim_{h \to 0} h^2 + 3 x^2 + 3xh + 7h + 14x + 8\]
\[ \Rightarrow f'(x) = 3 x^2 + 14x + 8\]
\[ \Rightarrow f'(x) = \lim_{h \to 0} \frac{(x + h )^3 + 7(x + h )^2 + 8(x + h) - 9 - x^3 - 7 x^2 - 8x + 9}{h}\]
\[ \Rightarrow f'(x) = \lim_{h \to 0} \frac{x^3 + h^3 + 3 x^2 h + 3x h^2 + 7 x^2 + 7 h^2 + 14xh + 8x + 8h - 9 - x^3 - 7 x^2 - 8x + 9}{h}\]
\[ \Rightarrow f'(x) = \lim_{h \to 0} \frac{h^3 + 3 x^2 h + 3x h^2 + 7 h^2 + 14xh + 8h}{h}\]
\[ \Rightarrow f'(x) = \lim_{h \to 0} \frac{h( h^2 + 3 x^2 + 3xh + 7h + 14x + 8)}{h}\]
\[ \Rightarrow f'(x) = \lim_{h \to 0} h^2 + 3 x^2 + 3xh + 7h + 14x + 8\]
\[ \Rightarrow f'(x) = 3 x^2 + 14x + 8\]
Thus,
\[f'(4) = 3 \times 4^2 + 14 \times 4 + 8 \]
\[ = 48 + 56 + 8\]
\[ = 112\]
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