Moderate -0.3 This is a straightforward application of the first principles definition to a polynomial. While it requires algebraic manipulation of the difference quotient and factoring (particularly for the cubic term), it follows a standard template with no conceptual surprises. The polynomial structure makes the algebra routine, placing it slightly below average difficulty.
6 Let \(\mathrm { f } ( x ) = 2 x ^ { 3 } + 3 x\). Use differentiation from first principles to show that \(\mathrm { f } ^ { \prime } ( x ) = 6 x ^ { 2 } + 3\).
Correct expansion of \((x+h)^3\). Must have numerical coefficients not \(^3C_1\). Condone 1 as a coefficient. Could use \(\delta x\) instead of \(h\). Allow unsimplified
Attempt to simplify \(f(x+h) - f(x)\). If considering \(2x^3\) and \(3x\) separately then both must be considered for M1. \(f(x+h)\) must be a 4 term cubic
\(= 6x^2h + 6xh^2 + 2h^3 + 3h\)
A1
Correct 4 term expression for \(f(x+h) - f(x)\) www
Attempt \(\frac{f(x+h)-f(x)}{h}\). f must be in terms of the given function. \(f(x+h)\) does not need to be expanded. Allow even if \(f(x+h)\) is now incorrect
Complete proof by considering limit as \(h \to 0\). Must see 'lim', '\(h \to 0\)', and \(f'(x)\). Dep on previous 5 marks. NB Starting with \(6x^2 + 3\) will get no credit
[6]
## Question 6:
| Answer | Mark | Guidance |
|--------|------|----------|
| $(x+h)^3 = x^3 + 3x^2h + 3xh^2 + h^3$ | B1 | Correct expansion of $(x+h)^3$. Must have numerical coefficients not $^3C_1$. Condone 1 as a coefficient. Could use $\delta x$ instead of $h$. Allow unsimplified |
| $f(x+h) - f(x) = \{2(x+h)^3 + 3(x+h)\} - \{2x^3 + 3x\}$ $= 2(x^3 + 3x^2h + 3xh^2 + h^3) + 3(x+h) - 2x^3 - 3x$ | M1 | Attempt to simplify $f(x+h) - f(x)$. If considering $2x^3$ and $3x$ separately then both must be considered for M1. $f(x+h)$ must be a 4 term cubic |
| $= 6x^2h + 6xh^2 + 2h^3 + 3h$ | A1 | Correct 4 term expression for $f(x+h) - f(x)$ www |
| $\frac{f(x+h)-f(x)}{h} = \frac{6x^2h + 6xh^2 + 2h^3 + 3h}{h}$ | M1 | Attempt $\frac{f(x+h)-f(x)}{h}$. f must be in terms of the given function. $f(x+h)$ does not need to be expanded. Allow even if $f(x+h)$ is now incorrect |
| $6x^2 + 6xh + 2h^2 + 3$ | A1 | Obtain correct expression www |
| $f'(x) = \lim_{h \to 0}(6x^2 + 6xh + 2h^2 + 3) = 6x^2 + 3$ | A1 | Complete proof by considering limit as $h \to 0$. Must see 'lim', '$h \to 0$', and $f'(x)$. Dep on previous 5 marks. **NB** Starting with $6x^2 + 3$ will get no credit |
| **[6]** | | |
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6 Let $\mathrm { f } ( x ) = 2 x ^ { 3 } + 3 x$. Use differentiation from first principles to show that $\mathrm { f } ^ { \prime } ( x ) = 6 x ^ { 2 } + 3$.
\hfill \mbox{\textit{OCR H240/01 2019 Q6 [6]}}