Answer to Question #277777 in Calculus for n hasan

Question

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

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 to a curve is a line perpendicular to a tangent to the curve.

equation of tangent:

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

$y=\sqrt{16-4x^2}$

$y'=-4x/\sqrt{16-4x^2}$

$y'(-1)=2/\sqrt 3$

$y-3=2(x+1)/\sqrt 3$

$y=2x/\sqrt 3+2/\sqrt 3+3$

equation of normal:

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

$y-3=-\sqrt 3(x+1)/2$

$2y=-x\sqrt 3+6-\sqrt 3$

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^∗

$f'(x) = 3x ^ 2 - 6x + 3x =0$

$x=\frac{2\pm \sqrt{4-4}}{2}=1$

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