Answer :
Sure, let's find the critical numbers of each function step-by-step.
### a. [tex]\( f(x) = x^3 + 6x^2 - 15x \)[/tex]
1. Find the first derivative [tex]\( f'(x) \)[/tex]:
[tex]\[ f'(x) = \frac{d}{dx}(x^3 + 6x^2 - 15x) = 3x^2 + 12x - 15 \][/tex]
2. Set the first derivative equal to zero and solve for [tex]\( x \)[/tex]:
[tex]\[ 3x^2 + 12x - 15 = 0 \][/tex]
3. Solve the quadratic equation:
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
where [tex]\( a = 3 \)[/tex], [tex]\( b = 12 \)[/tex], and [tex]\( c = -15 \)[/tex].
[tex]\[ x = \frac{-12 \pm \sqrt{12^2 - 4 \cdot 3 \cdot (-15)}}{2 \cdot 3} \][/tex]
[tex]\[ x = \frac{-12 \pm \sqrt{144 + 180}}{6} \][/tex]
[tex]\[ x = \frac{-12 \pm \sqrt{324}}{6} \][/tex]
[tex]\[ x = \frac{-12 \pm 18}{6} \][/tex]
[tex]\[ x = \frac{6}{6} = 1 \quad \text{or} \quad x = \frac{-30}{6} = -5 \][/tex]
Thus, the critical numbers are [tex]\( x = -5 \)[/tex] and [tex]\( x = 1 \)[/tex].
### b. [tex]\( f(x) = \frac{x}{x^2 - 9} \)[/tex]
1. Find the first derivative [tex]\( f'(x) \)[/tex]:
[tex]\[ f'(x) = \frac{d}{dx} \left( \frac{x}{x^2 - 9} \right) \][/tex]
Using the quotient rule [tex]\( \left( \frac{u}{v} \right)' = \frac{u'v - uv'}{v^2} \)[/tex]:
[tex]\[ u = x, \quad u' = 1, \quad v = x^2 - 9, \quad v' = 2x \][/tex]
[tex]\[ f'(x) = \frac{(1)(x^2 - 9) - (x)(2x)}{(x^2 - 9)^2} = \frac{x^2 - 9 - 2x^2}{(x^2 - 9)^2} = \frac{-x^2 - 9}{(x^2 - 9)^2} \][/tex]
[tex]\[ f'(x) = \frac{-x^2 - 9}{(x^2 - 9)^2} \][/tex]
2. Set the first derivative equal to zero and solve for [tex]\( x \)[/tex]:
[tex]\[ \frac{-x^2 - 9}{(x^2 - 9)^2} = 0 \][/tex]
Equating the numerator to zero:
[tex]\[ -x^2 - 9 = 0 \][/tex]
[tex]\[ -x^2 = 9 \][/tex]
[tex]\[ x^2 = -9 \][/tex]
[tex]\[ x = \pm 3i \][/tex]
Thus, the critical numbers are [tex]\( x = 3i \)[/tex] and [tex]\( x = -3i \)[/tex].
### c. [tex]\( f(x) = x^{\frac{4}{5}}(x - 4)^2 \)[/tex]
1. Find the first derivative [tex]\( f'(x) \)[/tex]:
Using the product rule [tex]\( (uv)' = u'v + uv' \)[/tex]:
[tex]\[ u = x^{\frac{4}{5}}, \quad v = (x - 4)^2 \][/tex]
[tex]\[ u' = \frac{4}{5}x^{-\frac{1}{5}}, \quad v' = 2(x - 4) \][/tex]
[tex]\[ f'(x) = \left( \frac{4}{5} x^{-\frac{1}{5}} \right) (x - 4)^2 + \left( x^{\frac{4}{5}} \right) \left( 2(x - 4) \right) \][/tex]
Simplify the expression:
[tex]\[ f'(x) = \frac{4}{5} x^{-\frac{1}{5}} (x - 4)^2 + 2x^{\frac{4}{5}} (x - 4) \][/tex]
2. Set the first derivative equal to zero and solve for [tex]\( x \)[/tex]:
[tex]\[ \frac{4}{5} x^{-\frac{1}{5}} (x - 4)^2 + 2 x^{\frac{4}{5}} (x - 4) = 0 \][/tex]
Factor out [tex]\( x^{-\frac{1}{5}} (x - 4) \)[/tex]:
[tex]\[ x^{-\frac{1}{5}} (x - 4) \left( \frac{4}{5} (x - 4) + 2 x \right) = 0 \][/tex]
Set each factor to zero:
[tex]\[ x^{-\frac{1}{5}} = 0 \quad \text{(no solution)}, \][/tex]
[tex]\[ x - 4 = 0 \implies x = 4, \][/tex]
[tex]\[ \frac{4}{5} (x - 4) + 2x = 0 \][/tex]
[tex]\[ \frac{4}{5} x - \frac{16}{5} + 2x = 0 \][/tex]
[tex]\[ \frac{4}{5} x + \frac{10}{5} x = \frac{16}{5} \][/tex]
[tex]\[ \frac{14}{5} x = \frac{16}{5} \][/tex]
[tex]\[ x = \frac{16}{14} = \frac{8}{7} \][/tex]
Thus, the critical numbers are [tex]\( x \approx 1.142857 \)[/tex] and [tex]\( x = 4 \)[/tex].
### Summary
- For [tex]\( f(x) = x^3 + 6x^2 - 15x \)[/tex]: The critical numbers are [tex]\( x = -5 \)[/tex] and [tex]\( x = 1 \)[/tex].
- For [tex]\( f(x) = \frac{x}{x^2 - 9} \)[/tex]: The critical numbers are [tex]\( x = 3i \)[/tex] and [tex]\( x = -3i \)[/tex].
- For [tex]\( f(x) = x^{\frac{4}{5}}(x - 4)^2 \)[/tex]: The critical numbers are [tex]\( x \approx 1.142857 \)[/tex] and [tex]\( x = 4 \)[/tex].
### a. [tex]\( f(x) = x^3 + 6x^2 - 15x \)[/tex]
1. Find the first derivative [tex]\( f'(x) \)[/tex]:
[tex]\[ f'(x) = \frac{d}{dx}(x^3 + 6x^2 - 15x) = 3x^2 + 12x - 15 \][/tex]
2. Set the first derivative equal to zero and solve for [tex]\( x \)[/tex]:
[tex]\[ 3x^2 + 12x - 15 = 0 \][/tex]
3. Solve the quadratic equation:
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
where [tex]\( a = 3 \)[/tex], [tex]\( b = 12 \)[/tex], and [tex]\( c = -15 \)[/tex].
[tex]\[ x = \frac{-12 \pm \sqrt{12^2 - 4 \cdot 3 \cdot (-15)}}{2 \cdot 3} \][/tex]
[tex]\[ x = \frac{-12 \pm \sqrt{144 + 180}}{6} \][/tex]
[tex]\[ x = \frac{-12 \pm \sqrt{324}}{6} \][/tex]
[tex]\[ x = \frac{-12 \pm 18}{6} \][/tex]
[tex]\[ x = \frac{6}{6} = 1 \quad \text{or} \quad x = \frac{-30}{6} = -5 \][/tex]
Thus, the critical numbers are [tex]\( x = -5 \)[/tex] and [tex]\( x = 1 \)[/tex].
### b. [tex]\( f(x) = \frac{x}{x^2 - 9} \)[/tex]
1. Find the first derivative [tex]\( f'(x) \)[/tex]:
[tex]\[ f'(x) = \frac{d}{dx} \left( \frac{x}{x^2 - 9} \right) \][/tex]
Using the quotient rule [tex]\( \left( \frac{u}{v} \right)' = \frac{u'v - uv'}{v^2} \)[/tex]:
[tex]\[ u = x, \quad u' = 1, \quad v = x^2 - 9, \quad v' = 2x \][/tex]
[tex]\[ f'(x) = \frac{(1)(x^2 - 9) - (x)(2x)}{(x^2 - 9)^2} = \frac{x^2 - 9 - 2x^2}{(x^2 - 9)^2} = \frac{-x^2 - 9}{(x^2 - 9)^2} \][/tex]
[tex]\[ f'(x) = \frac{-x^2 - 9}{(x^2 - 9)^2} \][/tex]
2. Set the first derivative equal to zero and solve for [tex]\( x \)[/tex]:
[tex]\[ \frac{-x^2 - 9}{(x^2 - 9)^2} = 0 \][/tex]
Equating the numerator to zero:
[tex]\[ -x^2 - 9 = 0 \][/tex]
[tex]\[ -x^2 = 9 \][/tex]
[tex]\[ x^2 = -9 \][/tex]
[tex]\[ x = \pm 3i \][/tex]
Thus, the critical numbers are [tex]\( x = 3i \)[/tex] and [tex]\( x = -3i \)[/tex].
### c. [tex]\( f(x) = x^{\frac{4}{5}}(x - 4)^2 \)[/tex]
1. Find the first derivative [tex]\( f'(x) \)[/tex]:
Using the product rule [tex]\( (uv)' = u'v + uv' \)[/tex]:
[tex]\[ u = x^{\frac{4}{5}}, \quad v = (x - 4)^2 \][/tex]
[tex]\[ u' = \frac{4}{5}x^{-\frac{1}{5}}, \quad v' = 2(x - 4) \][/tex]
[tex]\[ f'(x) = \left( \frac{4}{5} x^{-\frac{1}{5}} \right) (x - 4)^2 + \left( x^{\frac{4}{5}} \right) \left( 2(x - 4) \right) \][/tex]
Simplify the expression:
[tex]\[ f'(x) = \frac{4}{5} x^{-\frac{1}{5}} (x - 4)^2 + 2x^{\frac{4}{5}} (x - 4) \][/tex]
2. Set the first derivative equal to zero and solve for [tex]\( x \)[/tex]:
[tex]\[ \frac{4}{5} x^{-\frac{1}{5}} (x - 4)^2 + 2 x^{\frac{4}{5}} (x - 4) = 0 \][/tex]
Factor out [tex]\( x^{-\frac{1}{5}} (x - 4) \)[/tex]:
[tex]\[ x^{-\frac{1}{5}} (x - 4) \left( \frac{4}{5} (x - 4) + 2 x \right) = 0 \][/tex]
Set each factor to zero:
[tex]\[ x^{-\frac{1}{5}} = 0 \quad \text{(no solution)}, \][/tex]
[tex]\[ x - 4 = 0 \implies x = 4, \][/tex]
[tex]\[ \frac{4}{5} (x - 4) + 2x = 0 \][/tex]
[tex]\[ \frac{4}{5} x - \frac{16}{5} + 2x = 0 \][/tex]
[tex]\[ \frac{4}{5} x + \frac{10}{5} x = \frac{16}{5} \][/tex]
[tex]\[ \frac{14}{5} x = \frac{16}{5} \][/tex]
[tex]\[ x = \frac{16}{14} = \frac{8}{7} \][/tex]
Thus, the critical numbers are [tex]\( x \approx 1.142857 \)[/tex] and [tex]\( x = 4 \)[/tex].
### Summary
- For [tex]\( f(x) = x^3 + 6x^2 - 15x \)[/tex]: The critical numbers are [tex]\( x = -5 \)[/tex] and [tex]\( x = 1 \)[/tex].
- For [tex]\( f(x) = \frac{x}{x^2 - 9} \)[/tex]: The critical numbers are [tex]\( x = 3i \)[/tex] and [tex]\( x = -3i \)[/tex].
- For [tex]\( f(x) = x^{\frac{4}{5}}(x - 4)^2 \)[/tex]: The critical numbers are [tex]\( x \approx 1.142857 \)[/tex] and [tex]\( x = 4 \)[/tex].
