Question

7. The curve y = x3 + 2x2 – 3x has a gradient of 4 at points A and B.
Find the x-coordinates of A and B.

Answer 1

Answer:

The curve y = x3 + 2x2 – 3x has a gradient of 4 at points A and B.

Find the x-coordinates of A and B.

Step-by-step explanation:

the curve y = x3 + 2x2 – 3x has a gradient of 4 at points a and b.

find the x-coordinates of a and b.

Answer 2

$\large\underline{\sf{Solution-}}$

Given curve is

$\rm :\longmapsto\:y = {x}^{3} + {2x}^{2} - 3x - - - - (1)$

On differentiating both sides w. r. t. x, we get

$\rm :\longmapsto\:\dfrac{d}{dx}y =\dfrac{d}{dx}( {x}^{3} + {2x}^{2} - 3x)$

$\rm :\longmapsto\:\dfrac{dy}{dx} =\dfrac{d}{dx}{x}^{3} + 2\dfrac{d}{dx}{x}^{2} - 3\dfrac{d}{dx}x$

$\rm :\longmapsto\:\dfrac{dy}{dx} = {3x}^{2} + 4x - 3$

We know gradient or slope of a curve is

$\rm :\longmapsto\:Gradient \: of \: curve = \dfrac{dy}{dx}$

And it is given that

$\rm :\longmapsto\:\dfrac{dy}{dx} = 4$

$\rm :\longmapsto\: {3x}^{2} + 4x - 3 = 4$

$\rm :\longmapsto\: {3x}^{2} + 4x - 3 - 4 = 0$

$\rm :\longmapsto\: {3x}^{2} + 4x - 7= 0$

$\rm :\longmapsto\: {3x}^{2} + 7x - 3x - 7= 0$

$\rm :\longmapsto\:x(3x + 7) - 1(3x + 7) = 0$

$\rm :\longmapsto\:(3x + 7)(x - 1) = 0$

$\rm :\longmapsto\:x = 1 \: \: or \: \: x = - \dfrac{7}{3}$

Hᴇɴᴄᴇ,

➢ Coordinates of A and Coordinates of B on the given curve are shown in the below table.

$\begin{gathered}\boxed{\begin{array}{c|c} \bf x & \bf y = {x}^{3} + 2{x}^{2} - 3x \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad \qquad}{} \\ \sf 1 & \sf 0 \\ \\ \sf - \dfrac{7}{3} & \sf - \dfrac{140}{27} \end{array}} \\ \end{gathered}$

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$\begin{gathered}\boxed{\begin{array}{c|c} \bf f(x) & \bf \dfrac{d}{dx}f(x) \\ \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf k & \sf 0 \\ \\ \sf sinx & \sf cosx \\ \\ \sf cosx & \sf - \: sinx \\ \\ \sf tanx & \sf {sec}^{2}x \\ \\ \sf cotx & \sf - {cosec}^{2}x \\ \\ \sf secx & \sf secx \: tanx\\ \\ \sf cosecx & \sf - \: cosecx \: cotx\\ \\ \sf \sqrt{x} & \sf \dfrac{1}{2 \sqrt{x} } \\ \\ \sf logx & \sf \dfrac{1}{x}\\ \\ \sf {e}^{x} & \sf {e}^{x} \end{array}} \\ \end{gathered}$