Question #072cb

2 Answers
May 17, 2017

To find the slope, evaluate f'(0):

f'(0) = -2

Explanation:

The equation for the slope of the tangent line is:

f'(x) = 5x^4+ 3x^2-2

We can find local maxima by computing the next derivative, setting that equal to 0, and then solving for the value(s) of x.

Compute the next derivative:

f''(x) = 20x^3+ 6x

Set it equal to 0:

20x^3+ 6x = 0#

Factor:

2x(10x^2+3) = 0

x = 0 and x = +-sqrt(3/10)i

Discard the imaginary roots.

x = 0

Compute the next derivative:

f'''(x) = 60x^2+ 6

Evaluate it at x = 0:

f'''(0) = 60(0)^2+ 6 = 6

6> 0, therefore, the slope is a local minimum at x =0

To find the slope, evaluate f'(0):

f'(0) = -2

May 17, 2017

-2

Explanation:

The tangent line will be the derivative of the function.
f'(x)=5x^4+3x^2-2

To find the minimum value of this line we have to find the critical points. So we find the second derivative
f''(x)=20x^3+6x

The minimum will be when this function is equal to zero
0=20x^3+6x
0=x(20x^2+6)

From this we can see that either factor will only be equal to zero when x=0.

So we plug this into the tangent line formula to get the slope at this point.
f'(0)=5(0)^4+3(0)^2-2
f'(0)=0+0-2
f'(0)=-2