Question

17 mark for review selected values of the twice - differentiable functions f and g and their derivatives are given in the table above. the value of lim(x→2) (x²f(x) - 16)/(g(x) - 2) is a - 28 b - 12 c 28 d nonexistent

Explanation:

Step1: Check form of limit

When \(x
ightarrow2\), the numerator \(x^{2}f(x)-16\) becomes \(2^{2}f(2)-16 = 4\times4 - 16=0\), and the denominator \(g(x)-2\) becomes \(g(2)-2=2 - 2 = 0\). So, we can use L - H rule.

Step2: Differentiate numerator and denominator

By the product rule \((uv)^\prime=u^\prime v+uv^\prime\), the derivative of the numerator \(u = x^{2}\), \(v = f(x)\), so \((x^{2}f(x))^\prime=2xf(x)+x^{2}f^\prime(x)\), and the derivative of the denominator is \(g^\prime(x)\). Then our limit becomes \(\lim_{x
ightarrow2}\frac{2xf(x)+x^{2}f^\prime(x)}{g^\prime(x)}\).

Step3: Substitute \(x = 2\)

Substitute \(x = 2\) into \(\frac{2xf(x)+x^{2}f^\prime(x)}{g^\prime(x)}\). We know \(f(2)=4\), \(f^\prime(2)=3\) and \(g^\prime(2)=1\). Then \(\frac{2\times2\times4+2^{2}\times3}{1}=\frac{16 + 12}{1}=28\).

Answer:

C. 28