Questions: Question 1 f(x)=x^3-5x^2+7x-35 exact zeros y-intercept end behavior domain range relative maximum(s) relative minimum(s) Factored form

Solution Steps

We can factor the given function $f(x) = x^3 - 5x^2 + 7x - 35$ by grouping: $f(x) = x^2(x - 5) + 7(x - 5)$ $f(x) = (x^2 + 7)(x - 5)$

Setting $f(x) = 0$, we get $(x^2 + 7)(x - 5) = 0$. This gives us $x - 5 = 0$, so $x = 5$. And $x^2 + 7 = 0$, so $x^2 = -7$, which means $x = \pm i\sqrt{7}$.

The y-intercept is found by setting $x = 0$: $f(0) = (0)^3 - 5(0)^2 + 7(0) - 35 = -35$.

Since the leading term is $x^3$ with a positive coefficient, as $x \to \infty$, $f(x) \to \infty$ and as $x \to -\infty$, $f(x) \to -\infty$.

The domain of a polynomial function is all real numbers.

Since this is a cubic function with no restrictions, the range is all real numbers.

Taking the derivative of $f(x)$, we get $f'(x) = 3x^2 - 10x + 7$. Setting $f'(x) = 0$: $3x^2 - 10x + 7 = 0$ $(3x - 7)(x - 1) = 0$ $x = 1$ or $x = 7/3$

For $x=1$, $f(1) = (1-5)(1+7) = -4(8) = -32$. For $x=7/3$, $f(7/3) = (49/9 + 7)(7/3 - 5) = (49/9+63/9)(-8/3) = (112/9)(-8/3) = -896/27 \approx -33.19$.

Since the function goes from increasing to decreasing at $x=1$, it is a relative maximum. Since the function goes from decreasing to increasing at $x=7/3$, it is a relative minimum.

Final Answer