Question

answer the following true or false: let $f(x) = x^2 - 7x + 3$. the slope of the secant line through $(3, -9)$ and the point $(b, f(b))$ will eventually become equal to the slope of the tangent line to $y = f(x)$ at $x = 3$ when $b$ gets close enough to 3. true false

Explanation:

Step1: Recall the definition of the derivative

The slope of the tangent line to a function \( y = f(x) \) at a point \( x = a \) is given by the derivative \( f'(a) \). The slope of the secant line through \( (a, f(a)) \) and \( (b, f(b)) \) is given by \( \frac{f(b)-f(a)}{b - a} \). As \( b \) approaches \( a \), the slope of the secant line approaches the slope of the tangent line at \( x = a \), which is the definition of the derivative: \( f'(a)=\lim_{b
ightarrow a}\frac{f(b)-f(a)}{b - a} \).

Step2: Apply the definition to the given function

For the function \( f(x)=x^{2}-7x + 3 \) and the point \( a = 3 \), as \( b \) gets close to \( 3 \), the slope of the secant line through \( (3,f(3))=(3,- 9) \) (since \( f(3)=3^{2}-7\times3 + 3=9 - 21 + 3=-9 \)) and \( (b,f(b)) \) is \( \frac{f(b)-f(3)}{b - 3}=\frac{(b^{2}-7b + 3)-(-9)}{b - 3}=\frac{b^{2}-7b+12}{b - 3} \). Factoring the numerator: \( b^{2}-7b + 12=(b - 3)(b - 4) \), so \( \frac{(b - 3)(b - 4)}{b - 3}=b - 4 \) (for \( b
eq3 \)). As \( b
ightarrow3 \), \( b - 4
ightarrow3 - 4=-1 \), which is the slope of the tangent line at \( x = 3 \) (we can also find \( f'(x)=2x-7 \), so \( f'(3)=2\times3-7=6 - 7=-1 \)). So the statement that the slope of the secant line through \( (3,-9) \) and \( (b,f(b)) \) will eventually become equal to the slope of the tangent line at \( x = 3 \) as \( b \) gets close to \( 3 \) is True, because as \( b \) approaches \( 3 \), the slope of the secant line approaches the slope of the tangent line (by the definition of the derivative).

Answer:

True