Answer to Question #236230 in Trigonometry for luka

Question #236230

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


1
Expert's answer
2021-09-14T06:04:59-0400
f(x)=limh0f(x+h)f(x)hf'(x)=\lim\limits_{h\to 0}\dfrac{f(x+h)-f(x)}{h}


=limh04(x+h)+2(x+h)24x2x2h=\lim\limits_{h\to 0}\dfrac{4(x+h)+2(x+h)^2-4x-2x^2}{h}

=limh04x+4h+2x2+4xh+2h24x2x2h=\lim\limits_{h\to 0}\dfrac{4x+4h+2x^2+4xh+2h^2-4x-2x^2}{h}

=limh04h+4xh+2h2h=limh0(4+4x+2h)=\lim\limits_{h\to 0}\dfrac{4h+4xh+2h^2}{h}=\lim\limits_{h\to 0}(4+4x+2h)


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

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


f(a)=5f'(a)=5

4+4a=54+4a=5


4a=14a=1

a=14a=\dfrac{1}{4}



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