   Chapter 9.4, Problem 31E ### 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 31-34, find the coordinates of points where the graph of   f ( x ) has horizontal tangents. As a check, graph   f ( x ) and see whether the points you found look as though they have horizontal tangents.   f ( x ) = − x 3 + 9 x 2 − 15 x + 6

To determine

To calculate: The coordinates of point where the curve y=x3+9x215x+6 has horizontal tangents. Also check the solution by plotting the graph of the curve.

Explanation

Given Information:

The curve is given by the function y=x3+9x215x+6.

Formula Used:

The condition for horizontal tangent is

f(x)=0

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 of derivatives,

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

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:

The provided function is y=x3+9x215x+6.

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(x3)dx+d(9x2)dxd(15x)dx+d(6)dx

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

y=d(x3)dx+d(9x2)dxd(15x)dx+d(6)dxy=(3x31)+9(2x21)15(x11)+0y=3x2+18x15

The condition for horizontal tangent is f(x)=0

Thus, for the given curve,

3x2+18x15=03(x26x+5)=0

Simplify the calculation by splitting the middle term as

3(x25xx+5)=03(x5)(x1)=0

The above equation gives two values of x for f(x)=0

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