`f(x) = -x^3 + 7x^2 - 15x` Find all relative extrema, use the second derivative test where applicable.

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Chapter 3, 3.4 - Problem 34 - Calculus of a Single Variable (10th Edition, Ron Larson).
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jmfj | (Level 1) Adjunct Educator

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I)  Find the first derivative ->  f'(x) = -3x^2 + 14x - 15 and set it equal to zero to solve.  This will find the x-values of your "critical points". A critical point is where the graph humps. Solving, you get x = 5/3 and x = 3, meaning there is a hump (relative extrema) at those values. Because we don't/can't graph the function, we don't know if those humps are hills or valleys.  If necessary, you can find the exact point of those critical points by plugging 5/3 into the original function to determine the y-value (and likewise for x = 3).

II) Find the second derivative -> f''(x) = -6x + 14.  The second derivative will help you find the exact point on a graph where a graph changes from a hill to a valley (or vise versa).  This concept is called "concavity." Plug the critical points from Part One into the second derivative:  f''(5/3) = a positive number -- the number four to be exact, but that won't matter to you -- and f''(3) = a negative number -- again, the number negative four to be exact, but that won't matter to you.  Why does the exact value not matter to you, you ask?  If the second derivative gives you a positive number, then your graph is "concave up" (or CUP UP or a VALLEY) and if the second derivative gives you a negative number, then your graph is "concave down" (or CUP DOWN or a HILL).  [A cup that is up would make a valley, but a cup that is upside down would make a hill]. 

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loves2learn | (Level 3) Salutatorian

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Find the first derivative.

Given,

`f(x)=-x^3+7x^2-15x`

`f'(x)=-3x^2+14x-15 `

Now we need to find the value(s) of x that make the first derivative zero to find the critical points.

`0=-3x^2+14x-15 `

`0=(-3x+5)(x-3) `

`x=5/3 ` and `x=3 ` (critical points)

These will be a relative extrema if it changes sign, so find 2 values around both to test using the first derivative test, like 1 and 2 for the first  and 2 and 3 for the second.

`f'(1)=-4 `

`f'(2)=1 `

Therefore, this is a relative extrema. Find the y-value, which is -10.19. `(5/3,-10.19) `

Do the same for the other critical point

`f'(2)=1 `

This is also a relative extrema. Find the y-value, which is -9. `(3,-9) `

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