Finding a Curve from Conditions | CAIE 9709 Pure Mathematics 1 Notes

Pure Mathematics 1

Mathematics9709

Put each condition into the right equation

A question may leave a constant in the gradient rule as well as the constant introduced by integration. They are different unknowns. A height belongs in $y$ ; a tangent gradient belongs in $dy/dx$ .

A stationary point at $x=p$ means $f'(p)=0$ .
A non-zero normal gradient $m_n$ gives tangent gradient $-1/m_n$ , since perpendicular finite gradients multiply to −1.
Two points give two equations in the integrated function. Subtracting the equations can eliminate the added constant.

Worked example

Find the curve with dy/dx = 6x + k and normal gradient 1/2 at (1, −3).

The tangent gradient is −2, so use the derivative to find $k$ first.

\begin{gathered}6(1)+k=-2\\k=-8\\y=3x^2-8x+c\\-3=3-8+c\\c=2\end{gathered}

Answer

y=3x^2-8x+2

If the given normal is a line equation, first rearrange it into $y=mx+d$ to read its gradient. A vertical normal instead means a horizontal tangent; do not take a reciprocal of an undefined gradient.

Use all the given conditions

Your turnPractice

A function has $f'(x)=kx+2$ , $f(0)=3$ and $f(2)=11$ . Find $f(x)$ .

Show worked answer

\begin{aligned}f(x)&=\frac{k}{2}x^2+2x+c\\3&=c\\11&=2k+4+3\\k&=2\end{aligned}

Answer

f(x)=x^2+2x+3

The derivative and both supplied function values agree.

A stated maximum or minimum value is a height, not an $x$ -coordinate. Find the stationary input from $f'(x)=0$ , then use the stated height in $f(x)$ .

Your turnPractice

Given $f'(x)=8-2x$ and a maximum value of 20, find $f(x)$ .

Show worked answer

The derivative is zero at $x=4$ . Integration gives $f(x)=8x-x^2+c$ ; $20=16+c$ .

Answer

f(x)=8x-x^2+4

$f''(x)=-2<0$ confirms a maximum.

Put each condition into the right equation

A question may leave a constant in the gradient rule as well as the constant introduced by integration. They are different unknowns. A height belongs in $y$ ; a tangent gradient belongs in $dy/dx$ .

A stationary point at $x=p$ means $f'(p)=0$ .
A non-zero normal gradient $m_n$ gives tangent gradient $-1/m_n$ , since perpendicular finite gradients multiply to −1.
Two points give two equations in the integrated function. Subtracting the equations can eliminate the added constant.

Worked example

Find the curve with dy/dx = 6x + k and normal gradient 1/2 at (1, −3).

The tangent gradient is −2, so use the derivative to find $k$ first.

\begin{gathered}6(1)+k=-2\\k=-8\\y=3x^2-8x+c\\-3=3-8+c\\c=2\end{gathered}

Answer

y=3x^2-8x+2

If the given normal is a line equation, first rearrange it into $y=mx+d$ to read its gradient. A vertical normal instead means a horizontal tangent; do not take a reciprocal of an undefined gradient.

Use all the given conditions

Your turnPractice

A function has $f'(x)=kx+2$ , $f(0)=3$ and $f(2)=11$ . Find $f(x)$ .

Show worked answer

\begin{aligned}f(x)&=\frac{k}{2}x^2+2x+c\\3&=c\\11&=2k+4+3\\k&=2\end{aligned}

Answer

f(x)=x^2+2x+3

The derivative and both supplied function values agree.

A stated maximum or minimum value is a height, not an $x$ -coordinate. Find the stationary input from $f'(x)=0$ , then use the stated height in $f(x)$ .

Your turnPractice

Given $f'(x)=8-2x$ and a maximum value of 20, find $f(x)$ .

Show worked answer

The derivative is zero at $x=4$ . Integration gives $f(x)=8x-x^2+c$ ; $20=16+c$ .

Answer

f(x)=8x-x^2+4

$f''(x)=-2<0$ confirms a maximum.