Jtdylktuy

Is my answer correct?

f(x) = 2x² ← this is the parabola

f(3) = 2 * 9 = 18 → the parabola passes through A (3 ; 18), so its tangent line too

f'(x) = 4x ← this is the derivative

…and the derivative is the slope of the tangent line to the curve at x

f'(3) = 4 * 3 = 12 ← this is the slope of the tangent line to the curve at x = 3

Equation of the tangent line

The typical equation of a line is: y = mx + b → where m: slope and where b: y-intercept

You know that the slope of the tangent line is 12.

The equation of the tangent line becomes: y = 12x + b

The tangent line passes through A (3 ; 18), so these coordinates must verify the equation of the tangent line.

y = 12x + b

b = y - 12x → you substitute x and y by the coordinates of the point A (3 ; 18)

b = 18 - 36 = - 18

→ The equation of the tangent line is: y = 12x - 18

Intersection between the tangent line to the curve and the x-axis: → when y = 0

y = 12x - 18 → when y = 0

12x - 18 = 0

12x = 18

x = 3/2

→ Point B (3/2 ; 0)

Intersection between the vertical line passes through the point A and the x-axis: → when x = 3

→ Point C (3 ; 0)

The equation of the vertical line is: x = 3

Area of the region bounded by the parabola y = 2x², the tangent line to this parabola at (3 ; 18), and the x-axis.

= (area of the region bounded by the parabola y = 2x² and the x-axis) - (area of the triangle ABC)

= [∫ (from 0 to 3) of the parabola] - [(xC - xB).(yA - yC)/2]

= [∫ (from 0 to 3) 2x².dx] - [(xC - xB).(yA - yC)/2]

= { [(2/3).x³] from 0 to 3 } - { [3 - (3/2)].(18 - 0)/2 }

= [(2/3) * 3³] - { [(6/2) - (3/2)] * 9 }

= [(2/3) * 27] - { [(3/2) * 9 }

= 18 - (27/2)

= (36/2) - (27/2)

= 9/2 square unit

$y=2x^2$ has derivative $y'=4x$. So the derivative will be positive for all $x>0$. This means the tangent line will intersect the x-axis at some point $x_a<x$ for a given x.

This means the area under the parabola and bounded by the tangent line will have two separate regions, A region with $x<x_a$ which is made up of only area under the parabola, and a region $x>x_a$ where area is from the area between the parabola and the tangent to $(3,18)$.

The equation of a tangent line at $(x_0,y_0)$ is $$frac{y-f(x_0)}{x-x_0}=f'(x_0)$$

Or : $y=f'(x_0)(x-x_0)+f(x_0)$

The x intercept happens where $y=0$.

Requiring $y=0$ implies an x intercept of $x_c=frac{-f(x_0)}{f'(x_0)}+x_0$

So from the above arguments with $x_0=3$:

$$A=int_0^{x_c}2x^2dx+int_{x_c}^32x^2-(12x-18)dx$$

But this can be simplified. The area under the tangent line is a triangle. So the the parabola can be integrated ignoring the tangent line, and then subtracting the area of the triangle.

From the above, we know $(x_0-x_c)=frac{f(x_0)}{f'(x_0)}$, the base of the triangle. The height is just $f(x_0)$.

So the area of the triangle is $A_t=frac{1}{2}frac{f(x_0)^2}{f'(x_0)}$

So:

$$A=int_0^{x_0}f(x)dx-frac{1}{2}frac{f(x_0)^2}{f'(x_0)}$$

and solve for $x_0=3$.

In this form, an expression can be found for $x_0$ which extremizes the area.