C1 2017 7 Edexcel

7

(a) At x = 4 we have:

\begin{align*}f'(x) &= 30 + \frac {6 - 5 \times 4^2} {\sqrt 4} \\ &= 30 + \frac {6 - 80} 2 \\ & = 30 - \frac {74} 2 \\ & = 30 - 37 \\ & = -7\end{align*}

So the gradient of the line is -7 and it passes through (4, -8) so:

y + 8 = -7(x - 4)

so:

y = -7x + 28 - 8 = -7x + 20

(b)

It helps to write f'(x) in the form:

f'(x) = 30 + 6x^{-1/2} - 5 x^{3/2}

So that:

\begin{align*}f(x) &= 30x + \frac {6x^{1/2}} {\frac 1 2} - \frac {5 x^{5/2}} {\frac 5 2} \\ & = 30x + 12 \sqrt x - 2x^{5/2} + C\end{align*}

We have:

\begin{align*}f(4) & = 30 \times 4 + 12 \sqrt 4 - 2 \times 4^{5/2} + C \\ & = 120 + 24 - 2 \times 2^5 + C \\ & = 144 - 64 + C\\ & = 80 + C\end{align*}

So:

80 + C = -8

giving:

C = -88

So:

f(x) = 30x + 12 \sqrt x - 2x^{5/2} - 88

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