Answer to Question #277988 in Calculus for Nitya

Question

3. a) Define tangent and normal of a curve with figure. Also find the equation of tangent and normal of the ellipse (x ^ 2)/4 + (y ^ 2)/16 = 1 at the point (- 1, 3) .

b) Explain maximum and minimum value of a function with graphically. Evaluate maximum and minimum value of the function f(x) = x ^ 3 - 3x ^ 2 + 3x + 1

Answer 1

Solution:

a) A tangent to a curve is a line that touches the curve at one point and has the same slope as the curve at that point. A normal curve is a line perpendicular to a tangent to the curve.

equation of tangent:

$$y-y_0=f'(x_0)(x-x_0)$$

$$y= \sqrt{16−4x ^2 ​}$$

$$y ′ =\frac{−4x}{\sqrt{16−4x ^2 ​} }$$

$$y'(-1)=\frac{2}{\sqrt 3}$$

$$y-3=\frac{2(x+1)}{\sqrt3}$$

$$y=\frac{2x}{\sqrt3}+\frac{2}{\sqrt3}+3$$

equation of normal:

$$y-y_o=\frac{-(x-x_o)}{f ′ (x_o )}$$

$$y-3=\frac{-\sqrt3(x+1)}{2}$$

$$2y=-x\sqrt3+6-\sqrt3$$

b) function f defined on a domain X has a global (or absolute) maximum point at x^∗, if f(x^∗) ≥ f(x) for all x in X. Similarly, the function has a global (or absolute) minimum point at x^∗, if f(x^∗) ≤ f(x) for all x in X

f is said to have a local (or relative) maximum point at the point x^∗if there exists some ε > 0 such that f(x^∗) ≥ f(x) for all x in X within distance ε of x^∗. Similarly, the function has a local minimum point at x^∗, if f(x^∗) ≤ f(x) for all x in X within distance ε of x^∗

$$f ^′ (x)=3x ^2 −6x+3x=0$$

$$x=\frac{2±\sqrt{4-4}}{2}=1$$

since f'(x) does not change sign at x = 1, there is no local extremum

so, since

$f(x)\to \infin$ for $x\to \infin$

and

$f(x)\to - \infin$ for $x\to -\infin$

then there is no global minima or maxima