Answer to Question #203444 in Calculus for jay

Question #203444

Given that 𝑓(π‘₯) = 3π‘₯ 2 βˆ’ 4π‘₯ + 7, use the definition of the derivative to find 𝑓 β€² (π‘₯)


1
Expert's answer
2021-06-07T11:51:10-0400

Solution:

𝑓(π‘₯)=3π‘₯2βˆ’4π‘₯+7𝑓(π‘₯) = 3π‘₯^2 βˆ’ 4π‘₯ + 7

By definition of derivatives:

fβ€²(x)=lim⁑hβ†’0f(x+h)βˆ’f(x)h=lim⁑hβ†’0(3(x+h)2βˆ’4(x+h)+7)βˆ’(3x2βˆ’4xβˆ’7)h=lim⁑hβ†’0(3(x2+2xh+h2)βˆ’4xβˆ’4h)βˆ’(3x2βˆ’4x)h=lim⁑hβ†’03x2+6xh+3h2βˆ’4xβˆ’4hβˆ’3x2+4xh=lim⁑hβ†’06xh+3h2βˆ’4hh=lim⁑hβ†’0(6x+3hβˆ’4)=6xβˆ’4\begin{aligned} f^{\prime}(x) &=\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h} \\ &=\lim _{h \rightarrow 0} \frac{\left(3(x+h)^{2}-4(x+h)+7\right)-\left(3 x^{2}-4 x-7\right)}{h} \\ &=\lim _{h \rightarrow 0} \frac{\left(3\left(x^{2}+2 x h+h^{2}\right)-4 x-4 h\right)-\left(3 x^{2}-4 x\right)}{h} \\ &=\lim _{h \rightarrow 0} \frac{3 x^{2}+6 x h+3 h^{2}-4 x-4 h-3 x^{2}+4 x}{h} \\ &=\lim _{h \rightarrow 0} \frac{6 x h+3 h^{2}-4 h}{h} \\ &=\lim _{h \rightarrow 0}(6 x+3 h-4) \\ &=6 x-4 \end{aligned}


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