Chapter 9.6, Problem 25E

### Mathematical Applications for the ...

12th Edition
Ronald J. Harshbarger + 1 other
ISBN: 9781337625340

Chapter
Section

### Mathematical Applications for the ...

12th Edition
Ronald J. Harshbarger + 1 other
ISBN: 9781337625340
Textbook Problem

# In Problems 25-28, write the equation of the line tangent to the graph of each function at the indicated point. As a check, graph both the function and the tangent line you found to see whether it looks correct. 25 .  y  =( x 2 − 3 x  + 3) 3  at  ( 2, 1 )

To determine

To calculate: The equation of line tangent to the function y=(x23x+3)3 at the point (2,1) and check by plotting the graph of the tangent line and the function.

Explanation

Given Information:

The function is y=(x2âˆ’3x+3)3 and the line is tangent at the point (2,1).

Formula used:

Power rule:

If y=un and u is differentiable function of x, then,

dydx=nunâˆ’1â‹…dudx

According to the property of differentiation, if a function is of the form f(x)=u(x)+v(x), then,

fâ€²(x)=uâ€²(x)+vâ€²(x)

According to the product rule, if f(x)=u(x)â‹…v(x), then,

fâ€²(x)=uâ€²(x)â‹…v(x)+vâ€²(x)â‹…u(x)

According to the point slope form, the equation of a line with slope m and point (x1,y1) is,

yâˆ’y1=m(xâˆ’x1)

Calculation:

Consider the provided function,

y=(x2âˆ’3x+3)3

The equation of line tangent to the equation is found using the point-slope form.

The slope of tangent is found by differentiating the function.

Now, consider (x2âˆ’3x+3) to be u,

y=u3

Differentiate both sides with respect to x as,

dydx=ddx(u3)

Simplify using the power rule,

dydx=3â‹…u3âˆ’1â‹…dudx=3u2dudx

Substitute (x2âˆ’3x+3) for u,

dydx=3(x2âˆ’3x+3)2ddx(x2âˆ’3x+3)=3(x2âˆ’3x+3)2(ddx(x2)âˆ’ddx(3x)+ddx

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