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A function $y= f\left ( x \right )$ has a second order derivative ${f}''\left ( x \right )= 6\left ( x-1 \right ).$

If its graph passes through the point $\left ( 2,1 \right )$ and at that point the tangent to the graph is $y= 3x-5,$

then the function is

Option 1)
$\left ( x+1 \right )^{3}$

Option 2)
$\left ( x-1 \right )^{3}$

Option 3)
$\left ( x-1 \right )^{2}$

Option 4)
$\left ( x+1 \right )^{2}$

Answers (1)

best_answer

As we have learned

Equation of the tangent -

To find the equation of the tangent we need either one slope + one point or two points.

$\therefore \:\:(y-y_{\circ})=m(x_{\circ }-y_{\circ })$

$or\:\:(y-y_{2})=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}(x-x_{2})$

- wherein

Where $(x_{\circ},y_{\circ})$ is the point on the curve and M = M_T slope of tangent.

Geometrical interpretation of dy / dx -

Slope of tangent line is $tan\theta$ where $\theta$ is the angle made by the line with the +ve direction of x axis.

$\therefore \:\:\frac{dy}{dx}=tan\theta$

-

Equation of tangent -

Equation of tangent from other point on the curve.

$Let\:\:\:y=mx+c\:\:and\:(h,k)$ is the point where it passes through.Then $K=mh+C$ therefore $C=K-mh$

$y=mx+K-mh$

- wherein

Where m is slope of tangent.

$f'' (x ) = 6 (x-1 )\\ \rightarrow f'(x)= 3x^2 -6x +C \\\rightarrow f(x)= x^3 - 3x^2 + Cx+D\\we \: \: have \: \: , 1 = 8-12 +2 C +D \\\Rightarrow 2C+D = 5 .....(1)\\Also , f'(x) = 3 (slope )\: \: at \: \: (2,1) \\ so 3 = 12 -12 + C \\ \Rightarrow C = 3 \\ So , D = -1 \\\therefore f(x)= x^3 - 3x^2 +3x -1 = (x-1)^3$

Option 1)

$\left ( x+1 \right )^{3}$

Option 2)

$\left ( x-1 \right )^{3}$

Option 3)

$\left ( x-1 \right )^{2}$

Option 4)

$\left ( x+1 \right )^{2}$

Posted by

Himanshu

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