Sunday, May 19, 2019

IGNOU MTE 01 2019 SOLVED ASSIGNMENT 2(a)

Consider
f(x) = {x^3} - 1
Now
f(x + h) = {\left( {x + h} \right)^3} - 1

At

x = {x_0}the derivative by first principles is calculated as

f'\left( {{x_0}} \right) = {\lim }\limits_{h \to 0} \frac{{f\left( {{x_0} + h} \right) - f\left( {{x_0}} \right)}}{h}


\Rightarrow f'\left( {{x_0}} \right) = {\lim }\limits_{h \to 0} \frac{{{{\left( {{x_0} + h} \right)}^3} - {x_0}^3}}{h}
\Rightarrow f'\left( {{x_0}} \right) = {\lim }\limits_{h \to 0} \frac{{{x_0}^3 + {h^3} + 3x_0^2h + 3{x_0}{h^2} - {x_0}^3}}{h}
\Rightarrow f'\left( {{x_0}} \right) = {\lim }\limits_{h \to 0} {h^2} + 3x_0^2 + 3{x_0}h
\Rightarrow f'\left( {{x_0}} \right) = 3x_0^2

Now again we have

\frac{{dy}}{{dx}} = 3x_0^2

Slope of tangent to the curve

f\left( x \right) = 3{\left( { - 2} \right)^2} = 12units

Slope of normal = -1/12 units

Equation of tangent to the curve f(x)

\left( {y + 9} \right) = 12\left( {x + 2} \right)

\Rightarrow \left( {y + 9} \right) = 12x + 24

\Rightarrow 12x - y - 15 = 0

Equation of normal to the curve f(x)

\left( {y + 9} \right) = - 1/12\left( {x + 2} \right)

12y + 108 = - x - 2

12y + x + 110 = 0.

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