Question

If f(x) = 4X + 2X^2, find f^!(x) from the first principle and hence find the value of a given that f^!(a) = 5

Answer 1

$$f'(x)=\lim\limits_{h\to 0}\dfrac{f(x+h)-f(x)}{h}$$

$$=\lim\limits_{h\to 0}\dfrac{4(x+h)+2(x+h)^2-4x-2x^2}{h}$$

$$=\lim\limits_{h\to 0}\dfrac{4x+4h+2x^2+4xh+2h^2-4x-2x^2}{h}$$

$$=\lim\limits_{h\to 0}\dfrac{4h+4xh+2h^2}{h}=\lim\limits_{h\to 0}(4+4x+2h)$$

$$=4+4x+0=4+4x$$

$$f'(x)=4+4x$$

$$f'(a)=5$$

$$4+4a=5$$

$$4a=1$$

$$a=\dfrac{1}{4}$$