   Chapter 9.4, Problem 30E ### Mathematical Applications for the ...

11th Edition
Ronald J. Harshbarger + 1 other
ISBN: 9781305108042

#### Solutions

Chapter
Section ### Mathematical Applications for the ...

11th Edition
Ronald J. Harshbarger + 1 other
ISBN: 9781305108042
Textbook Problem

# In Problems 27-30, write the equation of the tangent line to each curve at the indicated point. As a check, graph both the function and the tangent line.   f ( x ) = x 3 3 − 3 x 3   at  x = − 1

To determine

To calculate: The equation of the tangent to curve y=x333x3 at x=1, also draw the graph to check both the function and the tangent line.

Explanation

Given Information:

The curve is given by the function y=x333x3, and the point is x=1.

Formula Used:

According to sum rule of derivatives,

If f(x)=u(x)+v(x)

Then, f(x)=u(x)+v(x)

According to power rule,

If f(x)=xn, then f(x)=nxn1

According to constant function rule,

If f(x)=c, then f(x)=0.

Equation of tangent at any point (x1,y1) is given by

yy1=y(xx1)

Coefficient rule for a constant c is such that, if f(x)=cu(x), where u(x) is a differentiable function of x, then f(x)=cu(x).

Calculation:

Consider the provided function,

y=x333x3

Since, the slope of the tangent at any point of the curve is given by derivative of the function.

Therefore, to calculate slope of the tangent calculate the derivative of the function.

Apply the sum rule of derivatives,

y=d(x33)dxd(3x3)dx

Use the power rule, coefficient rule and constant function rule to find the derivative of the function,

y=d(x33)dxd(3x3)dx=13(3x31)3(3x31)=x2+9x4

To determine the slope of tangent at x=1,

Substitute, x=1 in y=x2+9x4

y(1)=(1)2+9(1)4=1+9=10

Also at x=1,

y=(1)333(1)3=13+3=83

Equation of tangent at any point (x1,y1) is given by

yy1=y(xx1)

Thus, to calculate the equation of tangent at x=1 use the above formula and substitute (1,83) for (x1,y1)

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