Question

estimate the instantaneous rate of change of the function f(x)=16x^2-25 at x=-7. round your answer to a whole number

Answer 1

i applied the difference quotient equation to the function and then I got 224, is that about right?

Answer 2

32x , at x=-7 ??

Answer 3

the limit of the difference quotient; as h to zero ....

Answer 4

\[\lim_{h\to~0}~\frac{(16(x+h)^2-25)-(16x^2-25)}{(x+h)-x}\]
\[\lim_{h\to~0}~\frac{16(x^2+2xh+h^2)-25-16x^2+25}{h}\]
\[\lim_{h\to~0}~\frac{\cancel{16x^2}+2(16)xh+16h^2-\cancel{25}-\cancel{16x^2}+\cancel{25}}{h}\]
\[\lim_{h\to~0}~\frac{2(16)xh+16h^2}{h}\]
\[\lim_{h\to~0}~\frac{\cancel h(2(16)x+16h}{\cancel h}\]
\[\lim_{h\to~0}~2(16)x+16h = 32x\]

Answer 5

or by the derivative rule:\[\frac d{dx}[kx^n+c]=k~nx^{n-1}+0\]