How do use the first derivative test to determine the local extrema #x^2-2x-3#?

Question

Emily Ammerman · Accepted Answer

There is a minima of f(x) at x= 1

#f(x)= x^2 -2x-3#, hence f'(x)= 2x-2. Since f(x) is defined for all values of x, the only critical point is given by 2x-2=0, or, x=1 Since divide the entire domain in two parts #(-oo,1) and (1,+oo)# Now consider any value of x in #(-oo,1)#, say, x=-1. For x= -1, f'(x) would equal to -2-2= -4. This means the slope of the curve is negative. At x=-1, the slope is 0 At any point in #(1, +oo), say x=2, f'(x) would be 4-2=2. This means the slope of the curve is positive. The above analysis of the slope of f(x) , shows that the slope of the curve changes from negative, to the left of x=1, to positive to its right. This implies that there is a minima at x=1

Cameron Alexander · Answer

undefined To use the first derivative test to determine the local extrema of the function $f(x) = x^2 - 2x - 3$: 1. Find the first derivative of the function $f'(x)$. 2. Determine the critical points by solving $f'(x) = 0$. 3. Determine the sign of $f'(x)$ in intervals between the critical points. 4. Use the signs of $f'(x)$ to determine the behavior of the function and identify local extrema. Let's proceed with the steps: 1. Find the first derivative: $$f'(x) = 2x - 2$$ 2. Determine the critical points: $$f'(x) = 0 \implies 2x - 2 = 0 \implies x = 1$$ 3. Determine the sign of $f'(x)$ in intervals: - When $x < 1$, choose $x = 0$, then $f'(0) = -2$, which means $f'(x) < 0$. - When $x > 1$, choose $x = 2$, then $f'(2) = 2$, which means $f'(x) > 0$. 4. Use the signs of $f'(x)$: - $f'(x) < 0$ to the left of $x = 1$ indicates a decreasing function. - $f'(x) > 0$ to the right of $x = 1$ indicates an increasing function. Hence, by the first derivative test, there is a local minimum at $x = 1$.