Question

Question #072cb

Answer 1

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`

Answer 2

`-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`